Arithmetic Progression Class 10th Mathematics Gujarat Board Solution

Class 10th Mathematics Gujarat Board Solution
Exercise 5.1
  1. a = 3, d = 2 Given a and d for the following A.P., find the following A.P.:…
  2. a = -3, d = -2 Given a and d for the following A.P., find the following A.P.:…
  3. a = 100, d = -7 Given a and d for the following A.P., find the following A.P.:…
  4. a = -100, d = 7 Given a and d for the following A.P., find the following A.P.:…
  5. a = 1000, d = -100 Given a and d for the following A.P., find the following…
  6. 5, -5, 5, -5, ... Determine if the following sequences represent an A.P.,…
  7. 2, 2, 2, 2, ... Determine if the following sequences represent an A.P.,…
  8. 1, 11, 111, 1111, ... Determine if the following sequences represent an A.P.,…
  9. 5, 15, 25, 35, 45, ... Determine if the following sequences represent an A.P.,…
  10. 17, 22, 27, 32, ... Determine if the following sequences represent an A.P.,…
  11. 101, 99, 97, 95, ... Determine if the following sequences represent an A.P.,…
  12. 201, 198, 195, 192, ... Determine if the following sequences represent an A.P.,…
  13. Natural numbers which are consecutive multiples of 5 in increasing order.…
  14. Natural numbers which are multiples of 3 or 5 in increasing order. Determine if…
  15. 2, 7, 12, 17, ... Find the nth term of the following A.P.'s:
  16. 200, 195, 190, 185, ... Find the nth term of the following A.P.'s:…
  17. 1000, 900, 800, ... Find the nth term of the following A.P.'s:
  18. 50, 100, 150, 200, ... Find the nth term of the following A.P.'s:…
  19. 1/2 , 3/2 , 5/2 , 7/2 , 9/2 , l Find the nth term of the following A.P.'s:…
  20. 1.1, 2.1, 3.1, 4.1, ... Find the nth term of the following A.P.'s:…
  21. 1.2, 2.3, 3.4, 4.5, ... Find the nth term of the following A.P.'s:…
  22. 5/3 7/3 , 3 , 11/3 , 13/3 , 5 l Find the nth term of the following A.P.'s:…
  23. T7 = 12, T12 = 72 Find A.P. if Tn, Tm are as given below:
  24. T2 = 1, T12 = -9 Find A.P. if Tn, Tm are as given below:
  25. In an A.P., T3 = 8, T10 = T6 + 20. Find the A.P.
  26. In an A.P. 5th term is 17 and 9th term exceeds 2nd term by 35. Find the A.P.…
  27. Can any term of A.P., 12, 17, 22, 27, ... be zero? Why?
  28. Can any term of A.P., 201, 197, 193, ... be 5? Why?
  29. Which term of A.P., 8, 11, 14, 17, .. is 272?
  30. Find the 10th term from end for A.P., 3, 6, 9, 12, ... 300.
  31. Find the 15th term from end for A.P., 10, 15, 20, 25, 30, ..., 1000.…
  32. If in an A.P., T7 = 18, T18 = 7, find T101.
  33. If in an A.P., Tm = n, Tn = m, prove d = -1.
Exercise 5.2
  1. 2, 6, 10, 14, ... up to 20 terms Find the sum of the first n terms of the…
  2. 5, 7, 9, 11, ... upto 30 terms Find the sum of the first n terms of the…
  3. —10, —12, —14, —16, ... up to 15 terms Find the sum of the first n terms of the…
  4. 1, 1.5, 2, 2.5, 3, ... upto 16 terms Find the sum of the first n terms of the…
  5. 1/3 , 4/3 , 7/3 , 10/3 , l up to 18 terms Find the sum of the first n terms of…
  6. 3 + 6 + 9 + ... + 300 Find the sums indicated below:
  7. 5 + 10 + 15 + …… + 100 Find the sums indicated below:
  8. 7 + 12 + 17 + 22 + …… + 102 Find the sums indicated below:
  9. (-100) + (-92) + (-84) + …… + 92 Find the sums indicated below:
  10. 25 + 21 + 17 + 13 + …… + (-51) Find the sums indicated below:
  11. a = 1, d = 2, find S10. For a given A.P. with
  12. a = 2, d = 3, find S30. For a given A.P. with
  13. S3 = 9, S7 = 49, find Sn and S10. For a given A.P. with
  14. T10 = 41, S10 = 320, find Tn, Sn. For a given A.P. with
  15. S10 = 50, a = 0.5, find d. For a given A.P. with
  16. S20 = 100, d = —2, find a. For a given A.P. with
  17. How many terms of A.P., 2, 7, 12, 17, ... add up to 990?
  18. The first term of finite A.P. is 5, the last term is 45 and the sum is 500. Find…
  19. If the first term and the last term of a finite A.P. are 5 and 95 respectively…
  20. The sum of first n terms of an A.P. is 5n — 2n^2 . Find the A.P. i.e. a and d.…
  21. Find the sum of all three-digit numbers divisible by 3.
  22. Find the sum of all odd numbers from 5 to 205.
  23. Which term of A.P. 121, 117, 113, ... is its first negative term? If it is the…
Exercise 5
  1. If Tn = 6n + 5, find Sn
  2. If Sn = n^2 + 2n, find Tn
  3. If the sum of the first n terms of A, P. 30, 27, 24, 21 … is 120, find the…
  4. Which term of A.P., 100, 97, 94, 91, will be its first —ve term?
  5. Find the sum of all 3 digit natural multiples of 6.
  6. The ratio of the sum of m terms to sum of n terms of an A.P. is m^2/n^2 . Find…
  7. Sum to first l, m, n terms of A.P. are p, q, r. Prove that p/(m-n) + q/m (n-l) +…
  8. The ratio of sum to n terms of two A.P. is 8n+1/7n+3 for every n ∈ N. Find the…
  9. Three numbers in A.P. have the sum 18 and the sum of their squares is 180. Find…
  10. In a potato race a bucket is placed at the starting point ft is 5 m away from…
  11. A ladder has rungs 25 cm apart. The rungs decrease uniformly from 60 cm at…
  12. A man purchased LCD TV for Rs. 32, 500. He paid Rs. 200 initially and…
  13. In an A.P, T1= 22, Tn = —11, Sn = 66, find n.
  14. In an A.P. a = 8, Tn= 33, Sn = 123, find d and n.
  15. If T3 = 8, T7 = 24, then T10 = _____A. —4 B. 28 C. 32 D. 36
  16. If Sn = 2n^2 + 3n, then d =_____A. 13 B. 4 C. 9 D. —2
  17. If the sum of the three consecutive terms of A.P. is 48 and the product of the…
  18. If a = 2 and d = 4, then S20 = _____A. 600 B. 800 C. 78 D. 80
  19. If 3 + 5 + 7 + 9 + ... upto n terms = 288, then n = ……A. 12 B. 15 C. 16 D. 17…
  20. Four numbers are in A.P. and their sum is 72 and the largest of them is twice…
  21. If S1= 2 + 4 + ... + 2n and S2 = 1 + 3 + ... + (2n — 1), then S1: S2 = …..A.…
  22. For A.P., Sn — 2Sn - 1 + Sn - 2 =…. (n 2)A. 2d B. d C. a D. a + d…
  23. If Sm = n and Sn = m then Sm + n = ……A. —(m + n) B. 0 C. m + n D. 2m — 2n…
  24. If T4 = 7 and T7 = 4, then T10 = …..A. 9 B. 11 C. —11 D. 1
  25. If 2k + 1, 13, 5k — 3 are three consecutive terms of A.P., then k = ……….A. 17…
  26. (1) + (1 + 1) + (1 + 1 + 1) + ... + (1 + 1 + 1 + ...n — 1 times) =A. (n-1) n/2…
  27. In the A.P., 5, 7, 9, 11, 13, 15, ... the sixth term which is prime isA. 13 B.…
  28. For A.P. T18 — T8 = ………A. d B. 10d C. 26d D. 2d
  29. If for A.P., T25 — T20 = 15 then d =A. 3 B. 5 C. 20 D. 25

Exercise 5.1
Question 1.

Given a and d for the following A.P., find the following A.P.:

a = 3, d = 2


Answer:

Formula Used.


an = a + (n–1)d


If a = 3, d = 2


Then a1 = a + (n–1)d = 3 + (1–1)2 = 3


a2 = a + (n–1)d = 3 + (2–1)2 = 3 + 2×1 = 5


a3 = a + (n–1)d = 3 + (3–1)2 = 3 + 2×2 = 7


a4 = a + (n–1)d = 3 + (4–1)2 = 3 + 2×3 = 9


an = a + (n–1)d = 3 + (n–1)2 = 2n + 1


∴ The A.P is 3, 5, 7, 9……, 2n + 1



Question 2.

Given a and d for the following A.P., find the following A.P.:

a = –3, d = –2


Answer:

Formula Used.


an = a + (n–1)d


If a = –3, d = –2


Then a1 = a + (n–1)d = –3 + (1–1)(–2) = –3


a2 = a + (n–1)d = –3 + (2–1)(–2) = –3 + 1×(–2) = –5


a3 = a + (n–1)d = –3 + (3–1)(–2) = –3 + 2×(–2) = –7


a4 = a + (n–1)d = –3 + (4–1)(–2) = –3 + 3×(–2) = –9


an = a + (n–1)d = –3 + (n–1)(–2) = –1–2n


∴ The A.P is –3, –5, –7, –9……, –1–2n



Question 3.

Given a and d for the following A.P., find the following A.P.:

a = 100, d = –7


Answer:

Formula Used.


an = a + (n–1)d


If a = 100, d = –7


Then a1 = a + (n–1)d = 100 + (1–1)(–7) = 100


a2 = a + (n–1)d = 100 + (2–1)(–7) = 100 + (–7)×1 = 93


a3 = a + (n–1)d = 100 + (3–1)(–7) = 100 + (–7)×2 = 86


a4 = a + (n–1)d = 100 + (4–1)(–7) = 100 + (–7)×3 = 79


an = a + (n–1)d = 100 + (n–1)(–7) = 107 – 7n


∴ The A.P is 100, 93, 86, 79……107–7n



Question 4.

Given a and d for the following A.P., find the following A.P.:

a = –100, d = 7


Answer:

Formula Used.


an = a + (n–1)d


If a = –100, d = 7


Then a1 = a + (n–1)d = –100 + (1–1)7 = –100


a2 = a + (n–1)d = –100 + (2–1)7 = –100 + 7×1 = –93


a3 = a + (n–1)d = –100 + (3–1)7 = –100 + 7×2 = –86


a4 = a + (n–1)d = –100 + (4–1)7 = –100 + 7×3 = –79


an = a + (n–1)d = –100 + (n–1)7 = 7n–107


∴ The A.P is –100, –93, –86, –79……, 7n–107



Question 5.

Given a and d for the following A.P., find the following A.P.:

a = 1000, d = –100


Answer:

Formula Used.


an = a + (n–1)d


If a = 1000, d = –100


Then;


a1 = a + (n–1)d = 1000 + (1–1)(–100) = 1000


a2 = a + (n–1)d = 1000 + (2–1)(–100) = 1000 + (–100)×1 = 900


a3 = a + (n–1)d = 1000 + (3–1)(–100) = 1000 + (–100)×2 = 800


a4 = a + (n–1)d = 1000 + (4–1)(–100) = 1000 + (–100)×3 = 700


an = a + (n–1)d = 1000 + (n–1)(–100) = 1100–100n


∴ The A.P is 1000, 900, 800, 700……, 1100–100n



Question 6.

Determine if the following sequences represent an A.P., assuming that the pattern continues. If it is an A.P., find the nth term:

5, –5, 5, –5, ...


Answer:

Formula Used.


dn = an + 1 – an


In the above sequence,


a = 5;


d1 = a2–a1 = (–5)–5 = –10


d2 = a3–a2 = 5–(–5) = 10


d3 = a4–a3 = (–5)–5 = –10


⇒ As in A.P the difference between the 2 terms is always constant


But the difference in sequence is not the same.


∴ The above sequence is not A.P



Question 7.

Determine if the following sequences represent an A.P., assuming that the pattern continues. If it is an A.P., find the nth term:

2, 2, 2, 2, ...


Answer:

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = 2;


d1 = a2–a1 = 2–2 = 0


d2 = a3–a2 = 2–2 = 0


d3 = a4–a3 = 2–2 = 0


⇒ As in A.P the difference between the 2 terms is always constant


The difference in sequence comes to be 0.


∴ The above sequence is not an A.P



Question 8.

Determine if the following sequences represent an A.P., assuming that the pattern continues. If it is an A.P., find the nth term:

1, 11, 111, 1111, ...


Answer:

Formula Used.


dn = an + 1 – an


In the above sequence,


a = 1;


d1 = a2–a1 = 11–1 = 10


d2 = a3–a2 = 111–11 = 100


d3 = a4–a3 = 1111–111 = 1000


⇒ As in A.P the difference between the 2 terms is always constant


But the difference in sequence is not the same.


∴ The above sequence is not A.P



Question 9.

Determine if the following sequences represent an A.P., assuming that the pattern continues. If it is an A.P., find the nth term:

5, 15, 25, 35, 45, ...


Answer:

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = 5;


d1 = a2–a1 = 15–5 = 10


d2 = a3–a2 = 25–15 = 10


d3 = a4–a3 = 35–25 = 10


⇒ As in A.P the difference between the 2 terms is always constant


The difference in sequence is same and comes to be 10.


∴ The above sequence is A.P


The nth term of A.P is an = a + (n–1)d


an = a + (n–1)d = 5 + (n–1)10


= 5 + 10n–10


= –5 + 10n



Question 10.

Determine if the following sequences represent an A.P., assuming that the pattern continues. If it is an A.P., find the nth term:

17, 22, 27, 32, ...


Answer:

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = 17;


d1 = a2–a1 = 22–17 = 5


d2 = a3–a2 = 27–22 = 5


d3 = a4–a3 = 32–27 = 5


⇒ As in A.P the difference between the 2 terms is always constant


The difference in sequence is same and comes to be 5.


∴ The above sequence is A.P


The nth term of A.P is an = a + (n–1)d


an = a + (n–1)d = 17 + (n–1)5


= 17 + 5n–5


= 12 + 5n



Question 11.

Determine if the following sequences represent an A.P., assuming that the pattern continues. If it is an A.P., find the nth term:

101, 99, 97, 95, ...


Answer:

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = 101;


d1 = a2–a1 = 99–101 = –2


d2 = a3–a2 = 97–99 = –2


d3 = a4–a3 = 95–97 = –2


⇒ As in A.P the difference between the 2 terms is always constant


The difference in sequence is same and comes to be (–2).


∴ The above sequence is A.P


The nth term of A.P is an = a + (n–1)d


an = a + (n–1)d = 101 + (n–1)(–2)


= 101–2n + 2


= 103–2n



Question 12.

Determine if the following sequences represent an A.P., assuming that the pattern continues. If it is an A.P., find the nth term:

201, 198, 195, 192, ...


Answer:

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = 201;


d1 = a2–a1 = 198–201 = –3


d2 = a3–a2 = 195–198 = –3


d3 = a4–a3 = 192–195 = –3


⇒ As in A.P the difference between the 2 terms is always constant


The difference in sequence is same and comes to be (–3).


∴ The above sequence is A.P


The nth term of A.P is an = a + (n–1)d


an = a + (n–1)d = 201 + (n–1)(–3)


= 201–3n + 3


= 204–3n



Question 13.

Determine if the following sequences represent an A.P., assuming that the pattern continues. If it is an A.P., find the nth term:

Natural numbers which are consecutive multiples of 5 in increasing order.


Answer:

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


The sequence of natural numbers which are consecutive multiples of 5


Is 5, 10, 15, 20 ……


In the above sequence,


a = 5;


d1 = a2–a1 = 10–5 = 5


d2 = a3–a2 = 15–10 = 5


d3 = a4–a3 = 20–15 = 5


⇒ As in A.P the difference between the 2 terms is always constant


The difference in sequence is same and comes to be 5.


∴ The above sequence is A.P


The nth term of A.P is an = a + (n–1)d


an = a + (n–1)d = 5 + (n–1)5


= 5 + 5n–5


= 5n



Question 14.

Determine if the following sequences represent an A.P., assuming that the pattern continues. If it is an A.P., find the nth term:

Natural numbers which are multiples of 3 or 5 in increasing order.


Answer:

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


The sequence of natural numbers which are multiple of 3 or 5


in increasing order


Is 3, 5, 6, 9, 10 ……


In the above sequence,


a = 3;


d1 = a2–a1 = 5–3 = 2


d2 = a3–a2 = 6–5 = 1


d3 = a4–a3 = 9–6 = 3


⇒ As in A.P the difference between the 2 terms is always constant


The difference in sequence is not same .


∴ The above sequence is not A.P



Question 15.

Find the nth term of the following A.P.'s:

2, 7, 12, 17, ...


Answer:

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = 2;


d1 = a2–a1 = 7–2 = 5


d2 = a3–a2 = 12–7 = 5


d3 = a4–a3 = 17–12 = 5


The difference in sequence is same and comes to be 5 .


∴ The above sequence is A.P


The nth term of A.P is an = a + (n–1)d


an = a + (n–1)d = 2 + (n–1)(5)


= 2 + 5n–5


= –3 + 5n



Question 16.

Find the nth term of the following A.P.'s:

200, 195, 190, 185, ...


Answer:

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = 200;


d1 = a2–a1 = 195–200 = –5


d2 = a3–a2 = 190–195 = –5


d3 = a4–a3 = 185–190 = –5


The difference in sequence is same and comes to be (–5).


∴ The above sequence is A.P


The nth term of A.P is an = a + (n–1)d


an = a + (n–1)d = 200 + (n–1)(–5)


= 200–5n + 5


= 205–5n



Question 17.

Find the nth term of the following A.P.'s:

1000, 900, 800, ...


Answer:

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = 1000;


d1 = a2–a1 = 900–1000 = –100


d2 = a3–a2 = 800–900 = –100


The difference in sequence is same and comes to be (–100).


∴ The above sequence is A.P


The nth term of A.P is an = a + (n–1)d


an = a + (n–1)d = 1000 + (n–1)(–100)


= 1000–100n + 100


= 1100–100n



Question 18.

Find the nth term of the following A.P.'s:

50, 100, 150, 200, ...


Answer:

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = 50;


d1 = a2–a1 = 100–50 = 50


d2 = a3–a2 = 150–100 = 50


d3 = a4–a3 = 200–150 = 50


The difference in sequence is same and comes to be 50.


∴ The above sequence is A.P


The nth term of A.P is an = a + (n–1)d


an = a + (n–1)d = 50 + (n–1)(50)


= 50 + 50n–50


= 50n



Question 19.

Find the nth term of the following A.P.'s:



Answer:

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = ;


d1 = a2–a1 =  = 1


d2 = a3–a2 =  = 1


d3 = a4–a3 =  = 1


The difference in sequence is same and comes to be (1).


∴ The above sequence is A.P


The nth term of A.P is an = a + (n–1)d


an = a + (n–1)d =  + (n–1)(1)


 + n–1


= n–



Question 20.

Find the nth term of the following A.P.'s:

1.1, 2.1, 3.1, 4.1, ...


Answer:

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = 1.1;


d1 = a2–a1 = 2.1–1.1 = 1


d2 = a3–a2 = 3.1–2.1 = 1


d3 = a4–a3 = 4.1–3.1 = 1


The difference in sequence is same and comes to be (1).


∴ The above sequence is A.P


The nth term of A.P is an = a + (n–1)d


an = a + (n–1)d = 1.1 + (n–1)(1)


= 1.1 + n–1


= 0.1 + n



Question 21.

Find the nth term of the following A.P.'s:

1.2, 2.3, 3.4, 4.5, ...


Answer:

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = 1.2;


d1 = a2–a1 = 2.3–1.2 = 1.1


d2 = a3–a2 = 3.4–2.3 = 1.1


d3 = a4–a3 = 4.5–3.4 = 1.1


The difference in sequence is same and comes to be (1.1).


∴ The above sequence is A.P


The nth term of A.P is an = a + (n–1)d


an = a + (n–1)d = 1.2 + (n–1)(1.1)


= 1.2 + 1.1n–1.1


= 0.1 + 1.1n



Question 22.

Find the nth term of the following A.P.'s:



Answer:

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = ;


d1 = a2–a1 =  = 


d2 = a3–a2 =  = 


d3 = a4–a3 =  = 


The difference in sequence is same and comes to be ().


∴ The above sequence is A.P


The nth term of A.P is an = a + (n–1)d


an = a + (n–1)d =  + (n–1)()



= 1 + n



Question 23.

Find A.P. if Tn, Tm are as given below:

T7 = 12, T12 = 72


Answer:

Formula Used.


an = a + (n–1)d


a7 = a + (7–1)d


a7 = a + 6d


If 7th term of A.P is given as 12


Then,


a + 6d = 12


we get a = 12–6d ......eq 1


a12 = a + (12–1)d


a12 = a + 11d


If 12th term of A.P is given as 72


Then,


a + 11d = 72


we get a = 72–11d ......eq 2


Equating both eq 1 and eq 2


We get ;


12–6d = 72–11d


11d–6d = 72–12


5d = 60


d =  = 12


Putting d in eq 1 we get ;


a = 12–6×12


= 12–72 = –60


As a = –60 and d = 12


Then;


a1 = a + (n–1)d = –60 + (1–1)(12) = –60


a2 = a + (n–1)d = –60 + (2–1)(12) = –60 + (12)×1 = –48


a3 = a + (n–1)d = –60 + (3–1)(12) = –60 + (12)×2 = –36


a4 = a + (n–1)d = –60 + (4–1)(12) = –60 + (12)×3 = –24


an = a + (n–1)d = –60 + (n–1)(12) = 12n – 72


∴ The A.P is –60, –48, –36, –24……, 12n – 72



Question 24.

Find A.P. if Tn, Tm are as given below:

T2 = 1, T12 = –9


Answer:

Formula Used.


an = a + (n–1)d


a2 = a + (2–1)d


a2 = a + d


If 2nd term of A.P is given as 1


Then,


a + d = 1


we get a = 1–d ......eq 1


a12 = a + (12–1)d


a12 = a + 11d


If 12th term of A.P is given as –9


Then,


a + 11d = –9


we get a = –9–11d ......eq 2


Equating both eq 1 and eq 2


We get ;


1–d = –9–11d


11d–d = –9–1


10d = –10


d =  = –1


Putting d in eq 1 we get ;


a = 1–(–1)


= 2


As a = 2 and d = –1


Then;


a1 = a + (n–1)d = 2 + (1–1)(–1) = 2


a2 = a + (n–1)d = 2 + (2–1)(–1) = 2 + (–1)×1 = 2–1 = 1


a3 = a + (n–1)d = 2 + (3–1)(–1) = 2 + (–1)×2 = 2–2 = 0


a4 = a + (n–1)d = 2 + (4–1)(–1) = 2 + (–1)×3 = 2–3 = –1


an = a + (n–1)d = 2 + (n–1)(–1) = 2 + 1–n = 3–n


∴ The A.P is 2, 1, 0, –1……, 3–n



Question 25.

In an A.P., T3 = 8, T10 = T6 + 20. Find the A.P.


Answer:

Formula Used.


an = a + (n–1)d


a3 = a + (3–1)d


a3 = a + 2d


If 3rd term of A.P is given as 8


Then,


a + 2d = 8


we get a = 8–2d ......eq 1


a10 = a + (10–1)d


a10 = a + 9d


If 10th term of A.P is given as 20 + 6th term


Then,


a6 = a + (6–1)d


= a + 5d


a + 9d = 20 + [a + 5d]


we get


a–a + 9d–5d = 20


4d = 20


d =  = 5 ......eq 2


Putting d in eq 1 we get ;


a = 8–(2×5)


= 8–10 = –2


As a = –2 and d = 5


Then;


a1 = a + (n–1)d = –2 + (1–1)(5) = –2


a2 = a + (n–1)d = –2 + (2–1)(5) = –2 + (5)×1 = –2 + 5 = 3


a3 = a + (n–1)d = –2 + (3–1)(5) = –2 + (5)×2 = –2 + 10 = 8


a4 = a + (n–1)d = –2 + (4–1)(5) = –2 + (5)×3 = –2 + 15 = 13


an = a + (n–1)d = –2 + (n–1)(5) = –2–5 + 5n = –7 + 5n


∴ The A.P is –2, 3, 8, 13, ……, 5n–7



Question 26.

In an A.P. 5th term is 17 and 9th term exceeds 2nd term by 35. Find the A.P.


Answer:

Formula Used.


an = a + (n–1)d


a5 = a + (5–1)d


a5 = a + 4d


If 5th term of A.P is given as 17


Then,


a + 4d = 17


we get a = 17–4d ......eq 1


a9 = a + (9–1)d


a9 = a + 8d


If 9th term of A.P is given as 35 + 2nd term


Then,


a2 = a + (1)d


= a + d


a + 8d = 35 + [a + d]


we get


a–a + 8d–d = 35


7d = 35


d =  = 5 ......eq 2


Putting d in eq 1 we get ;


a = 17–(4×5)


= 17–20 = –3


As a = –3 and d = 5


Then;


a1 = a + (n–1)d = –3 + (1–1)(5) = –3


a2 = a + (n–1)d = –3 + (2–1)(5) = –3 + (5)×1 = –3 + 5 = 2


a3 = a + (n–1)d = –3 + (3–1)(5) = –3 + (5)×2 = –3 + 10 = 7


a4 = a + (n–1)d = –3 + (4–1)(5) = –3 + (5)×3 = –3 + 15 = 12


an = a + (n–1)d = –3 + (n–1)(5) = –3 + 5n–5 = 5n–8


∴ The A.P is –3, 2, 7, 12, ……, 5n–8



Question 27.

Can any term of A.P., 12, 17, 22, 27, ... be zero? Why?


Answer:

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = 12;


d1 = a2–a1 = 17–12 = 5


d2 = a3–a2 = 22–17 = 5


d3 = a4–a3 = 27–22 = 5


The difference in sequence is same and comes to be (5).


For any term of A.P to be 0


an = a + (n–1)d = 0


an = a + (n–1)d = 12 + (n–1)(5)


= 12 + 5n–5


= 7 + 5n


7 + 5n = 0


5n = –7


n = 


∴ The number of terms cannot be negative


Hence, the no term of A.P can be 0



Question 28.

Can any term of A.P., 201, 197, 193, ... be 5? Why?


Answer:

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = 201;


d1 = a2–a1 = 197–201 = –4


d2 = a3–a2 = 193–197 = –4


The difference in sequence is same and comes to be (–4).


For any term of A.P to be 5


an = a + (n–1)d = 5


an = a + (n–1)d = 201 + (n–1)(–4)


5 = 201–4n + 4


5 = 205–4n


–4n = –205 + 5


n =  = 50


∴ The 50th term of A.P is 5



Question 29.

Which term of A.P., 8, 11, 14, 17, .. is 272?


Answer:

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = 8;


d1 = a2–a1 = 11–8 = 3


d2 = a3–a2 = 14–11 = 3


d3 = a4–a3 = 17–14 = 3


The difference in sequence is same and comes to be (3).


For any term of A.P to be 272


an = a + (n–1)d = 272


an = a + (n–1)d = 8 + (n–1)(3)


= 8 + 3n–3


= 5 + 3n


5 + 3n = 272


3n = 272–5 = 267


n =  = 89


∴ The 89th term of A.P has value 272



Question 30.

Find the 10th term from end for A.P., 3, 6, 9, 12, ... 300.


Answer:

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = 3;


d1 = a2–a1 = 6–3 = 3


d2 = a3–a2 = 9–6 = 3


d3 = a4–a3 = 12–9 = 3


The difference in sequence is same and comes to be (3).


As the last term is 300


an = a + (n–1)d


300 = 3 + (n–1)3


300 = 3 + 3n–3 = 3n


n =  = 100th term


10th term from the end is


100 + 1–10 = 91st term


∵ while counting, last term is also counted


an = a + (n–1)d


a91 = 3 + (91–1)3


= 3 + 90×3


= 273



Question 31.

Find the 15th term from end for A.P., 10, 15, 20, 25, 30, ..., 1000.


Answer:

Formula Used.


dn = an + 1 – an


an = a + (n–1)d


In the above sequence,


a = 10;


d1 = a2–a1 = 15–10 = 5


d2 = a3–a2 = 20–15 = 5


d3 = a4–a3 = 25–20 = 5


The difference in sequence is same and comes to be (5).


As the last term is 1000


an = a + (n–1)d


1000 = 10 + (n–1)5


1000 = 10 + 5n–5 = 5 + 5n


5n = 995


n =  = 199th term


15 term from the end will be


199 + 1–15 = 185th term


an = a + (n–1)d


a185 = 10 + (185–1)5


= 10 + 184×5


= 10 + 920


= 930



Question 32.

If in an A.P., T7 = 18, T18 = 7, find T101.


Answer:

Formula Used.


an = a + (n–1)d


a7 = a + (7–1)d


a7 = a + 6d


If 7th term of A.P is given as 18


Then,


a + 6d = 18


we get a = 18–6d ......eq 1


a18 = a + (18–1)d


a18 = a + 17d


If 18th term of A.P is given as 7


Then,


a + 17d = 7


we get a = 7–17d ......eq 2


Equating both eq 1 and eq 2


We get ;


18–6d = 7–17d


17d–6d = 7–18


11d = –11


d =  = –1


Putting d in eq 1 we get ;


a = 18–6×(–1)


= 18 + 6 = 24


For 101st term


an = a + (n–1)d


a101 = 24 + (101–1)(–1)


= 24–100


= –76



Question 33.

If in an A.P., Tm = n, Tn = m, prove d = –1.


Answer:

Formula Used.


an = a + (n–1)d


am = a + (m–1)d


If mth term of A.P is given as n


Then,


a + (m–1)d = n


we get a = n–(m–1)d ......eq 1


an = a + (n–1)d


If nth term of A.P is given as m


Then,


a + (n–1)d = m


we get a = m–(n–1)d ......eq 2


Equating both eq 1 and eq 2


We get ;


n–(m–1)d = m–(n–1)d


(n–1)d–(m–1)d = m–n


(n–1–(m–1))d = m–n


(n–1–m + 1)d = m–n


(n–m)d = –(n–m)


d =  = –1




Exercise 5.2
Question 1.

Find the sum of the first n terms of the following A.P. as asked for:

2, 6, 10, 14, ... up to 20 terms


Answer:

Formula used.


Sn = [2a + (n–1)d]


d = an + 1–an


In the following A.P


a = 2,


d = a2–a1


= 6–2 = 4;


n = 20


Sn = [2a + (n–1)d]


[2×2 + (20–1)4]


= 10[4 + 19×4]


= 10[4 + 76]


= 800



Question 2.

Find the sum of the first n terms of the following A.P. as asked for:

5, 7, 9, 11, ... upto 30 terms


Answer:

Formula used.


Sn = [2a + (n–1)d]


d = an + 1–an


In the following A.P


a = 5,


d = a2–a1


= 7–5 = 2;


n = 30


Sn = [2a + (n–1)d]


[2×5 + (30–1)2]


= 15[10 + 29×2]


= 15[10 + 58]


= 1020



Question 3.

Find the sum of the first n terms of the following A.P. as asked for:

—10, —12, —14, —16, ... up to 15 terms


Answer:

Formula used.


Sn = [2a + (n–1)d]


d = an + 1–an


In the following A.P


a = –10,


d = a2–a1


= –12–(–10) = (–2);


n = 15


Sn = [2a + (n–1)d]


[2×(–10) + (15–1)(–2)]


[–20 + 14×(–2)]


[–20–28]


= 15×(–24)


= –360



Question 4.

Find the sum of the first n terms of the following A.P. as asked for:

1, 1.5, 2, 2.5, 3, ... upto 16 terms


Answer:

Formula used.


Sn = [2a + (n–1)d]


d = an + 1–an


In the following A.P


a = 1,


d = a2–a1


= 1.5–1 = 0.5;


n = 16


Sn = [2a + (n–1)d]


[2×1 + (16–1)0.5]


= 8[2 + 15×0.5]


= 8[2 + 7.5]


= 76



Question 5.

Find the sum of the first n terms of the following A.P. as asked for:

 up to 18 terms


Answer:

Formula used.


Sn = [2a + (n–1)d]


d = an + 1–an


In the following A.P


a = ,


d = a2–a1


 = 1;


n = 18


Sn = [2a + (n–1)d]


[2× + (18–1)1]


= 9[ + 17]


= 9[]


= 159



Question 6.

Find the sums indicated below:

3 + 6 + 9 + ... + 300


Answer:

Formula used.


Sn = [2a + (n–1)d]


d = an + 1–an


In the sum above A.P follows


Is 3, 6, 9, ……, 300


In the following A.P


⇒ a = 3,


⇒ d = 6–3 = 3


As the last term is 300


an = a + (n–1)d


300 = 3 + (n–1)3


300 = 3 + 3n–3


3n = 300


⇒ n =  = 100


Sn = [2a + (n–1)d]


[2×3 + (100–1)3]


= 50[6 + 99×3]


= 50[6 + 297]


= 15150



Question 7.

Find the sums indicated below:

5 + 10 + 15 + …… + 100


Answer:

Formula used.


Sn = [2a + (n–1)d]


d = an + 1–an


In the sum above A.P follows


Is 5, 10, 15, ……, 100


In the following A.P


⇒ a = 5,


⇒ d = 10–5 = 5


As the last term is 100


an = a + (n–1)d


100 = 5 + (n–1)5


100 = 5 + 5n–5


5n = 100


⇒ n =  = 20


Sn = [2a + (n–1)d]


[2×5 + (20–1)5]


= 10[10 + 19×5]


= 10[10 + 95]


= 1050



Question 8.

Find the sums indicated below:

7 + 12 + 17 + 22 + …… + 102


Answer:

Formula used.


Sn = [2a + (n–1)d]


d = an + 1–an


In the sum above A.P follows


Is 7, 12, 17, 22, ……, 102


In the following A.P


⇒ a = 7,


⇒ d = 12–7 = 5


As the last term is 102


an = a + (n–1)d


102 = 7 + (n–1)5


102 = 7 + 5n–5


5n = 102–2


⇒ n =  = 20


Sn = [2a + (n–1)d]


[2×7 + (20–1)5]


= 10[14 + 19×5]


= 10[14 + 95]


= 1090



Question 9.

Find the sums indicated below:

(–100) + (–92) + (–84) + …… + 92


Answer:

Formula used.


Sn = [2a + (n–1)d]


d = an + 1–an


In the sum above A.P follows


Is –100, –92, –84, ……, 92


In the following A.P


⇒ a = –100,


⇒ d = –92–(–100) = –92 + 100 = 8


As the last term is 92


an = a + (n–1)d


92 = –100 + (n–1)8


92 = –100 + 8n–8


8n = 92 + 108


⇒ n =  = 25


Sn = [2a + (n–1)d]


[2×(–100) + (25–1)8]


[–200 + 24×8]


[–200 + 192]


= –100



Question 10.

Find the sums indicated below:

25 + 21 + 17 + 13 + …… + (–51)


Answer:

Formula used.


Sn = [2a + (n–1)d]


d = an + 1–an


In the sum above A.P follows


Is 25, 21, 17, 13, ……, (–51)


In the following A.P


⇒ a = 25,


⇒ d = 21–25 = –4


As the last term is –51


an = a + (n–1)d


–51 = 25 + (n–1)(–4)


–51 = 25–4n + 4


4n = 80


⇒ n =  = 20


Sn = [2a + (n–1)d]


[2×25 + (20–1)(–4)]


= 10[50 + 19×(–4)]


= 10[50–76]


= –260



Question 11.

For a given A.P. with

a = 1, d = 2, find S10.


Answer:

Formula used.


Sn = [2a + (n–1)d]


d = an + 1–an


In the following A.P


⇒ a = 1,


⇒ d = 2


⇒ n = 10


Sn = [2a + (n–1)d]


[2×1 + (10–1)2]


= 5[2 + 9×2]


= 5[2 + 18]


= 100



Question 12.

For a given A.P. with

a = 2, d = 3, find S30.


Answer:

Formula used.


Sn = [2a + (n–1)d]


d = an + 1–an


In the following A.P


⇒ a = 2,


⇒ d = 3


⇒ n = 30


Sn = [2a + (n–1)d]


[2×2 + (30–1)3]


= 15[4 + 29×3]


= 15[4 + 87]


= 1365



Question 13.

For a given A.P. with

S3 = 9, S7 = 49, find Sn and S10.


Answer:

Formula used.


Sn = [2a + (n–1)d]


d = an + 1–an


In the following A.P


⇒ S3 = 9,


S3 = [2a + (3–1)d]


9 = [2a + 2d]


9 = [2(a + d)]


 = [a + d]


3 = [a + d]


⇒ a = 3–d ……eq 1


⇒ S7 = 49


S7 = [2a + (7–1)d]


49 = [2a + 6d]


49 = [2(a + 3d)]


 = a + 3d


7 = a + 3d


⇒ a = 7–3d ......eq 2


By equating eq 1 and eq 2


3–d = 7–3d


3d – d = 7 – 3


2d = 4


d =  = 2


Putting value of d in eq 1


a = 3–d = 3–2


= 1


Sn = [2a + (n–1)d]


[2×1 + (n–1)2]


[2 + 2n–2]


×2n


= n2


∴ Sn = n2


⇒ S10 = (10)2


= 100



Question 14.

For a given A.P. with

T10 = 41, S10 = 320, find Tn, Sn.


Answer:

Formula used.


Sn = [a + an]


Tn = an = a + (n–1)d


Sn = [2a + (n–1)d]


⇒ S10 = 320


⇒ T10 = a10 = 41


S10 = [a + 41]


320 = [a + 41]


 = a + 41


64 = a + 41


a = 64–41


⇒ a = 23


a10 = a + (n–1)d


41 = 23 + (10–1)d


41–23 = 9d


9d = 18


⇒ d =  = 2


Tn = an = a + (n–1)d


Tn = 23 + (n–1)2


23–2 + 2n


21 + 2n


Sn = [2a + (n–1)d]


Sn = [2×23 + (n–1)2]


[46 + 2n–2]


[44 + 2n]


= n[22 + n]


= 22n + n2



Question 15.

For a given A.P. with

S10 = 50, a = 0.5, find d.


Answer:

Formula used.


Sn = [2a + (n–1)d]


⇒ S10 = 50


⇒ n = 10


⇒ a = 0.5S10 = [2×0.5 + (10–1)d]


50 = 5[1 + 9d]


 = 1 + 9d


9d = 10–1


d =  = 1



Question 16.

For a given A.P. with

S20 = 100, d = —2, find a.


Answer:

Formula used.


Sn = [2a + (n–1)d]


⇒ S20 = 100


⇒ n = 20


⇒ d = –2


S20 = [2a + (20–1)(–2)]


100 = 10[2a–38]


100 = 20[a–19]


 = a–19


5 = a–19


a = 19 + 5


a = 24



Question 17.

How many terms of A.P., 2, 7, 12, 17, ... add up to 990?


Answer:

Formula used.


Sn = [2a + (n–1)d]


d = an + 1–an


⇒ S20 = 990


⇒ a = 2


⇒ d = an + 1–an


= 7 – 2 = 5


Sn = [2×2 + (n–1)5]


990 = [4 + 5n–5]


990 = [5n–1]


990×2 = 5n2–n


5n2–n–1980 = 0


n = 


∴ n can be 20 or –19.8


Since, the number of terms cannot be negative


∴ Sum of 20 terms gives the sum of 990



Question 18.

The first term of finite A.P. is 5, the last term is 45 and the sum is 500. Find the number of terms.


Answer:

Formula used.


Sn = [a + l]


⇒ Sn = 500


⇒ a = 5


⇒ l = 45


Sn = [a + l]


500 = [5 + 45]


500×2 = n×50


n =  = 20


∴ Number of terms to get sum 500 is 20



Question 19.

If the first term and the last term of a finite A.P. are 5 and 95 respectively and d = 5, find n and Sn.


Answer:

Formula used.


an = a + (n–1)d


Sn = [2a + (n–1)d]


⇒ d = 5


⇒ a = 5


⇒ l = 95


an = a + (n–1)d


95 = 5 + (n–1)5


95 = 5–5 + 5n


n =  = 19


Sn = [a + l]


[5 + 95]


= 19×50


= 950


∴ Number of terms is 19 and Sum of A.P is 950



Question 20.

The sum of first n terms of an A.P. is 5n — 2n2. Find the A.P. i.e. a and d.


Answer:

Formula used.


Sn = [2a + (n–1)d]


Sn = [2a + (n–1)d]


5n – 2n2 = [2a + (n–1)d]


2×n[5 – 2n] = n[2a + (n–1)d]


10 – 4n = (2a–d) + nd


Then


Real part should be equal to Real


And imaginary part should be equal to imaginary


2a – d = 10 ......eq 1


nd = –4n


⇒ d = –4


Putting value of d in eq 1


2a – d = 10


2a – (–4) = 10


2a + 4 = 10


2a = 10 – 4 = 6


⇒ a =  = 3



Question 21.

Find the sum of all three–digit numbers divisible by 3.


Answer:

Formula used.


Sn = [a + l]


an = a + (n–1)d


⇒ The 1st 3 digit number divisible by 3 is 102


For the last 3 digit number divisible by 3


Divide 999 by 3


It gets completely divisible by 3


⇒ The last 3 digit number divisible by 3 is 999


an = a + (n–1)d


999 = 102 + (n–1)3


999 = 102–3 + 3n


3n = 999 – 99


n =  = 300


Sn = [a + l]


[102 + 999]


= 150×1101


= 165150



Question 22.

Find the sum of all odd numbers from 5 to 205.


Answer:

Formula used.


Sn = [a + l]


an = a + (n–1)d


⇒ The 1st odd number is 5


⇒ The last odd number is 205


⇒ Difference between every odd number is 2


an = a + (n–1)d


205 = 5 + (n–1)2


205 = 5–2 + 2n


2n = 205 – 3


n =  = 101


Sn = [a + l]


[5 + 205]


= 101×105


= 10605



Question 23.

Which term of A.P. 121, 117, 113, ... is its first negative term? If it is the nth term, find Sn.


Answer:

Formula used.


Sn = [2a + (n–1)d]


an = a + (n–1)d


d = an + 1 – an


Let the 1st negative term be X


a = 121


d = 117 – 121


= –4


If the difference is –4 then


Then X can be (–1), (–2), (–3), (–4)


an = a + (n–1)d


X = 121 + (n–1)(–4)


X = 125–4n


4n = 125 – X


n = 


For the number of terms to be an integer (125–X) will be multiple of 4


When we put X = –1


n will be 


which is not an integer


When we put X = –2


n will be 


which is not an integer


When we put X = –3


n will be  = 32


which is an integer


Hence –3 is the 1st negative term.


Sn = [2a + (n–1)d]


[2×121 + (32–1)(–4)]


= 16[242 – 124]


= 16×118


= 1888




Exercise 5
Question 1.

If Tn = 6n + 5, find Sn


Answer:

From the question we know that, Tn = 6n + 5 ……. (1)


And we know that Tn = a + (n – 1)d


So we have,


⇒ a + (n – 1)d = 6n + 5


⇒ a + (n – 1)d = 6n + 11 – 6


⇒ a + (n – 1)d = 11 + (6n – 6)


⇒ a + (n – 1)d = 11 + 6(n – 1)


On comparing both the sides, we have:


⇒ a = 11 and d = 6


So now, Sn =  × (2a + (n – 1)d)


⇒ Sn =  × (2(11) + (n – 1)(6))


⇒ Sn =  × (22 + 6n – 6)


⇒ Sn =  × (16 + 6n)


⇒ Sn = n × (8 + 3n)


⇒ Sn = 3n2 + 8n


∴ Sum of n terms of the given A.P is Sn = 3n2 + 8n



Question 2.

If Sn = n2 + 2n, find Tn


Answer:

From the question we know that, Sn = n2 + 2n ……. (1)


We know that, Sn =  × (2a + (n – 1)d)


= na + 


= na + (n2 – n) × 


Sn = na + n2 ×  – n ×  ……. (2)


As we know that, (1) and (2) are both sum of the same arithmetic progression


So we can equate them to each other.


So, we have,


⇒ n2 + 2n = na + n2 ×  – n × 


⇒ n2 + 2n = n2 × + na – n × 


Now we will equate the coefficients of “n2” and “n” on both the sides of the equal to sign.


So, we get,


For coefficients of n2 :


⇒ 1 = 


⇒ d=2


Now, For coefficients of n:


⇒ 2 = a – 


From above we have that d=2,


So we can say that,


⇒ 2 = a – 


⇒ 2 = a – 1


⇒ a=3


So, now, we know that, Tn = a + (n – 1)d


Now we put values of a and n in the above equation,


We get,


⇒ Tn = 3 + (n – 1)2


⇒ Tn = 3 + 2n – 2


⇒ Tn = 2n + 1


∴ for the given value of Sn = n2 + 2n, we have Tn = 2n + 1.



Question 3.

If the sum of the first n terms of A, P.

30, 27, 24, 21 … is 120, find the number of terms and the last term.


Answer:

We can see that, A.P. is 30, 27, 24, 21 …


T1 = 30 = a


T2 = 27


So, we have d = T2 – T1


d = 27 – 30


d= – 3


Now, we have sum of the A.P. Sn =  × (2a + (n – 1)d)


And it is given that, Sn = 120


So, we have,


⇒ 120 =  × (2a + (n – 1)d)


Now we will put the values of a and d in the above equation


⇒ 120 =  × (2(30) + (n – 1)(– 3))


⇒ 120 =  × (60 + 3 – 3n)


120 × 2 = n × (63 – 3n)


⇒ 240 = 63n – 3n2


⇒ 3n2 – 63n + 240 = 0


⇒ n2 – 21n + 80 = 0


⇒ n2 – 16n – 5n + 80 = 0


⇒ n(n – 16) – 5(n – 16) = 0


⇒ (n – 16)(n – 5) = 0


⇒ n= 16 or 5


Now, when n=16, the last term = T16


So we have, T16 = a + (16 – 1)d ……. (∵ Tn = a + (n – 1)d )


⇒ T16 = a + 15d


Now we put the values of a and d in the above equation


⇒ T16= 30 + 15(– 3)


⇒ T16 = 30 – 45


⇒ T16 = – 15


Now, when n=5, the last term = T5


So we have, T5 = a + (5 – 1)d ……. (∵ Tn = a + (n – 1)d )


⇒ T5 = a + 4d


Now we put the values of a and d in the above equation


⇒ T5= 30 + 4(– 3)


⇒ T5 = 30 – 12


⇒ T5 = 18


∴ The no. Of term in the above A.P. can be n = 5 or 16


And the last terms of the respective A.Ps are:


For n = 5, last term T5 = 18


For n = 16, last term T16 = – 15



Question 4.

Which term of A.P., 100, 97, 94, 91, will be its first —ve term?


Answer:

We have the A.P as 100, 97, 94, 91 …


So, a = 100


⇒ d= 97 – 100 = – 3


Now recall that, nth term of an AP is Tn = a + (n – 1)d


⇒ Tn = 100 + (n – 1)(– 3)


= 100 – 3n + 3


⇒ Tn = 103 – 3n


Since we need negative term, Tn < 0


So, 103 – 3n < 0


⇒ 103 < 3n


⇒  < n


⇒ 34.333 < n


⇒ n>34


i.e. N = 35


∴ n=35 for the first negative term in the given A.P.


⇒ T35 = a + (35 – 1)d


= 100 + 34 × – 3


= 100 – 102


= – 2


∴ the first negative term of the given AP is – 2.



Question 5.

Find the sum of all 3 digit natural multiples of 6.


Answer:

Now the first 3 digit multiple of 6 is 102.


That means, a = 102


And all the nos. Are multiples of 6 which means that they have a difference of 6 in between them so, common difference d = 6


Now, the last 3 digit multiple of 6 is 996.


Now, we know that, Tn = a + (n – 1)d


So, 996 = a + (n – 1)d


⇒ 996 = 102 + (n – 1)6


⇒ 6(n – 1) = 996 – 102


⇒ 6(n – 1) = 894


⇒ n – 1 = 


⇒ n – 1 = 149


⇒ n = 150


So, we can say that there are 150, 3 digit multiples of 6


Now sum of these nos. S150 =  × (2a + (n – 1)d)


⇒ S150 =  × (2(102) + (150 – 1)6)


= 75 × (204 + 149 × 6)


= 75 × (1098)


⇒ S150 = 82350


∴ Sum of all 3 digit natural multiples of 6 is 82350.



Question 6.

The ratio of the sum of m terms to sum of n terms of an A.P. is. Find the ratio of its mth term to its nth term.


Answer:

As we know that sum of n terms of an A.P is:


⇒ Sn × (2a + (n – 1)d)


So, sum of m terms of an A.P is:


⇒ Sm =  × (2a + (m – 1)d)


From the question we know that,


⇒  = 


⇒  =


⇒  = 


⇒  = 


⇒ m × (2a + (n – 1)d) = n × (2a + (m – 1)d)


⇒ 2ma + m(n – 1)d = 2na + n(m – 1)d


⇒ 2ma + md(n – 1) = 2na + nd(m – 1)


⇒ 2ma + mnd – md = 2na + mnd – nd


⇒ 2ma – md = 2na – nd


⇒ 2ma – 2na = md – nd


⇒ 2a(m – n) = d(m – n)


⇒ 2a = d


Now, mth term of the given A.P is Tm = a + (m – 1)d


Now, nth term of the given A.P is Tn = a + (n – 1)d


Now ratio between these 2 terms is:


⇒  = 


Now we have, 2a = d


So we get,


⇒  = 


⇒  = 


⇒  = 


⇒  = 


⇒  = 


∴ the ratio of mth term of the given A.P to its nth Term is:


Tm : Tn = 2m – 1 : 2n – 1



Question 7.

Sum to first l, m, n terms of A.P. are p, q, r. Prove that 


Answer:

Let the first term of AP be a and common difference be d.


Sum of the first p terms is:


Sl =  × (2a + (l – 1)d) = p ………(1)


Sum of the first m terms is:


⇒ Sm =  × (2a + (m – 1)d) = q ……… (2)


Sum of the first n terms is:


⇒ Sn =  × (2a + (n – 1)d) = r ……… (3)


Now, (1) ×  + (2) ×  + (3) × 


We get, p ×  + q ×  + r × 




= 0 + [(m – n)l + (n – l)m + (l – m)n –m + n – n + l – l + m]


[lm – ln + mn – lm + ln – mn + 0]


= × 0 = 0


∴ it is proved that, p ×  + q ×  + r ×  = 0



Question 8.

The ratio of sum to n terms of two A.P. is for every n ∈ N. Find the ratio of their 7th terms and mth terms.


Answer:

Let the sum of n terms of the first A.P be:


⇒ Sn =  × (2a + (n – 1)d) …………… (1)


Let the sum of n terms of the second A.P be:


⇒ S’n =  × (2a’ + (n – 1)d’) …………… (2)


Now according to the question:


⇒  = 


Let’s consider the ratio these two AP’s mth terms as:


Tm : T’m


Now, recall that, nth term of an AP is Tn = a + (n – 1)d


⇒ Tm = a + (m – 1)d


⇒ T’m = a’ + (m – 1)d’


Hence the ratio of these two AP’s mth terms become:


⇒  = 


On multiplying by 2, we get,


⇒  = 


⇒  = 


⇒  = 


⇒  = 


⇒  = 


⇒  = 


Now from the above formula of the ratio of mth terms of 2 Aps, we can find the ratio of 7th terms of both Aps


So we have,  = 


⇒  = 


⇒  = 


∴ the ratio of mth terms of the given 2 Aps is, 16m – 7 : 14m – 4


∴ the ratio of 7th terms of the given 2 Aps is, 105 : 94



Question 9.

Three numbers in A.P. have the sum 18 and the sum of their squares is 180. Find the numbers in the increasing order.


Answer:

We know that the sum of 3 numbers in the AP is 18.


Let’s say that those 3 numbers are: a – d, a, a + d


So we can say that, a – d + a + a + d = 18


⇒ 3a = 18


⇒ a= 6


Now, we have that the sum of squares of these 3 numbers is 180.


So we can say that, (a – d) 2 + a2 + (a + d) 2 = 180


⇒ a2 – 2ad + d2 + a2 + a2 + 2ad + d2 = 180


⇒ 3a2 + 2d2 = 180


We know that a = 6,


So, 3(6)2 + 2d2 = 180


⇒ 3(36) + 2d2 = 180


⇒ 108 + 2d2 = 180


⇒ 2d2 = 180 – 108


⇒ 2d2 = 72


⇒ d2 = 36


⇒ d = 6


Now, our 3 numbers a – d, a, a + d are 6 – 6, 6, 6 + 6 = 0, 6, 12 respectively.


∴ The 3 numbers of the A.P in the increasing order are : 0, 6, 12.



Question 10.

In a potato race a bucket is placed at the starting point ft is 5 m away from the first potato. The rest of the potatoes are placed in a straight line each 3. In away from the other. Each competitor starts from the bucket. Picks up the nearest potato and runs back and drops it in the bucket and continues till all potatoes are placed in the bucket What is the total distance covered if 15 potatoes are placed in the race?



If the distance covered is 1340 m, find the number of potatoes?


Answer:

From the given data we calculate the distance covered for each potato.


So, for first potato, distance = 2 × 5 = 10m


For second, distance = 10 + 2 × 3= 16m


For third, distance = 16 + 2 × 3= 22m


Thus the distance to be covered form an AP. :


10, 16, 22, 28, …………


The total distance to be covered for 15 potatoes is given by S15 .


Hence, for the above AP, we have a = 10 and d = 6.


So, we know that, Sn =  × (2a + (n – 1)d)


⇒ S15= × [2(10) + (15 – 1) × 6]


 × [20 + 14 × 6]


 × [20 + 84]


 × 104


= 15 × 52


= 780m


∴ if 15 potatoes are placed in the race, the total distance covered is 780m.


Now, it is given that, the total distance covered is 1340m and we need to find the no. Of potatoes.


So, let the no. Of potatoes be n, then we take,


⇒ Sn = 1340


⇒ 1340 =  × (2a + (n – 1)d)


⇒ 1340 =  × (2(10) + (n – 1)(6))


⇒ 1340 =  × (20 + 6n – 6)


⇒ 1340 =  × (14 + 6n)


⇒ 1340 = n × (7 + 3n)


⇒ 3n2 + 7n – 1340 = 0


On solving we get that,


⇒ n =  or n = 20


as we n cannot be negative, so we have n = 20.


∴ if total distance to be covered is 1340m, then no. Of potatoes placed in the race are 20.



Question 11.

A ladder has rungs 25 cm apart. The rungs decrease uniformly from 60 cm at bottom to 40 cm at top. If the distance between the top rung and the bottom rung is 2.5 in, find the length of the wood required.



Answer:

The distance between 2 consecutive rungs is 25cm and the distance between the top and bottom rung is 2.5m = 250cm.


∴ no. Of rungs =  + 1 = 11


The length of the bottom rung is 60cm and going upwards the length of the rung decreases uniformly.


The length of last rung is 40cm.


So, the length of the rung will form a finite AP.


With first term a = 60(T1) and the last 11th term as 40(T11)


Now, d = 




⇒ d= – 2


the length of the wood required is given by S11


we know that, Sn =  × (2a + (n – 1)d)


So, S11 =  × (2a + (n – 1)d)


⇒ S11 =  × (2(60) + (11 – 1)( – 2))


⇒ S11 =  × ( 120 – 20)


⇒ S11 =  × 100


⇒ S11 = 550


∴ the length of the wood required is 550cm i.e. 5.5m.



Question 12.

A man purchased LCD TV for Rs. 32, 500. He paid Rs. 200 initially and increasing the payment by Rs. 150 every month. How many months did he take to make the complete payment?


Answer:

The mount paid as down payment is 200Rs.


Amount paid in 1st instalment = 200 + 150 = 350Rs


Amount paid in 2nd instalment = 350 + 150 = 500Rs and so on.


The amount paid every month increase every month and forms finite AP. Which is : 200, 350, 500 ……..


Total amount paid is Sn = 32, 500Rs


we know that, Sn =  × (2a + (n – 1)d)


⇒ 32, 500 =  × (2(200) + (n – 1)(150))


⇒ 32, 500 =  × (400 + 150n – 150)


⇒ 32, 500 =  × (250 + 150n)


⇒ 32, 500 =n × (125 + 75n)


⇒ 75n2 + 125n – 32500 = 0


⇒ 25n2 + 5n – 1300 = 0


⇒ (3n + 65)(n – 20)=0


⇒ n =  or n = 20


As n cannot be negative, so we have n = 20.


Thus we can say that there are 20 terms in the AP.


Among these terms first payment is the down payment, so man took 20 – 1 = 19 months to complete the payment.


∴ man took 19 months to complete the payment.



Question 13.

In an A.P, T1= 22, Tn = —11, Sn = 66, find n.


Answer:

We know that, Tn = a + (n – 1)d


So for T1,


We have T1=a + (1 – 1)d = a


According to the question, a = 22


Now, we have Tn = – 11


Ie., a + (n – 1)d = – 11


So, 22 + (n – 1)d = – 11


⇒ (n – 1)d = – 22 – 11


⇒ (n – 1)d = – 33


Now we have that, Sn =  × (2a + (n – 1)d)


From the question we can say that, Sn = 66


So, we have,


⇒ 66=  × (2a + (n – 1)d)


⇒ 132 = n × (2a + (n – 1)d)


We have (n – 1)d = – 33 and a = 22, so we put that in the above equation.


⇒ 132 = n × (2(22) – 33)


⇒ 132 = n × (44 – 33)


⇒ 132= n × 11


⇒ n = 


⇒ n = 12


∴ value of n is 12.



Question 14.

In an A.P. a = 8, Tn= 33, Sn = 123, find d and n.


Answer:

We have a = 8


We know that, Tn = a + (n – 1)d


And from the question we can say that, Tn= 53


So we get,


⇒ 33 = a + (n – 1)d


Putting value of a in the above equation,


⇒ 33 = 8 + (n – 1)d


⇒ (n – 1)d = 33 – 8


⇒ (n – 1)d = 25


Now we have that, Sn =  × (2a + (n – 1)d)


From the question we can say that, Sn = 123


So, we have,


⇒ 123 =  × (2a + (n – 1)d)


⇒ 246 = n × (2a + (n – 1)d)


now we have a = 8 and (n – 1)d = 25, so we put them in the above equation,


⇒ 246 = n × (2(8) + 25)


⇒ 246 = n × (16 + 25)


⇒ 246 = n × (31)


⇒ n=  = 6


⇒ n = 6


now, we know that, (n – 1)d = 25


we put the value of n in the above equation, we get


⇒ (6 – 1)d = 25


⇒ 5d = 25


⇒ d = 5


∴ for the given A.P, value of n is 6 and value of d is 5.



Question 15.

If T3 = 8, T7 = 24, then T10 = _____
A. —4

B. 28

C. 32

D. 36


Answer:

We have, T3 = 8


And recall that, Tn = a + (n – 1)d


So we have, T3 = a + (3 – 1)d


⇒ 8 = a + 2d ……(1)


We also have, T7 = 24


And T7 = a + (7 – 1)d


⇒ 24 = a + 6d …… (2)


From (1), we have that, a = 8 – 2d


So we put the value of a in (2)


⇒ 24 = 8 – 2d + 6d


⇒ 24 – 8 = 4d


⇒ 16 = 4d


⇒ d= 4


now, putting value of d in (1),


⇒ 8 = a + 2(4)


⇒ 8 = a + 8


⇒ a =0


now, T10= a + (10 – 1)d


⇒ T10= 0 + 9 × 4


⇒ T10= 36


∴ correct option is (d).


Question 16.

If Sn = 2n2 + 3n, then d =_____
A. 13

B. 4

C. 9

D. —2


Answer:

We have Sn = 2n2 + 3n from the question.


We know that, Sn =  × (2a + (n – 1)d)


So we can say that,


⇒ 2n2 + 3n =  × (2a + (n – 1)d)


⇒ 4n2 + 6n = n × (2a + (n – 1)d)


⇒ 4n2 + 6n = 2na + dn2 – dn


⇒ 4n2 + 6n = dn2 + (2a – d)n


Now comparing the coefficients of “n2” and “n”, we get that,


d = 4


∴ the correct option is (b).


Question 17.

If the sum of the three consecutive terms of A.P. is 48 and the product of the first and the last is 252, then d = _____
A. 2

B. 3

C. 4

D. 16


Answer:

We know that the sum of 3 nos in the AP is 48.


Lets say that those 3 nos are : a – d, a, a + d


So we can say that, a – d + a + a + d = 48


3a = 48


a= 16


now we have product of the first and the last term out of these 3 to be 252.


Ie. (a – d)(a + d) = 252


⇒ a2 – d2 = 252


we have a = 16, so we put the value of a in the above equation:


⇒ (16)2 – d2 = 252


⇒ 256 – d2 = 252


⇒ d2 = 256 – 252


⇒ d2 = 4


⇒ d=2


∴ the correct option is (a)


Question 18.

If a = 2 and d = 4, then S20 = _____
A. 600

B. 800

C. 78

D. 80


Answer:

We have a = 2


d= 4


n = 20


we also know that, Sn =  × (2a + (n – 1)d)


⇒ S20 =  × (2(2) + (20 – 1)4)


⇒ S20 = 10 × (4 + 19 × 4)


⇒ S20 = 10 × (4 + 76)


⇒ S20 = 10 × 80


⇒ S20 = 800


∴ the correct option is (b)


Question 19.

If 3 + 5 + 7 + 9 + ... upto n terms = 288, then n = ……
A. 12

B. 15

C. 16

D. 17


Answer:

We have a = 3


Then d = 5 – 3 = 2


Then, Sn = 288


We can recall that, Sn =  × (2a + (n – 1)d)


So, Sn =  × (2(3) + (n – 1)2)


⇒ Sn =  × (6 + 2n – 2)


⇒ Sn =  × (4 + 2n)


⇒ Sn = n × (2 + n)


⇒ Sn = n2 + 2n


⇒ n2 + 2n = 288


⇒ n2 + 2n – 288 = 0


⇒ n2 + 18n – 16n – 288 = 0


⇒ n(n + 18) – 16(n + 18) = 0


⇒ (n + 18)(n – 16)= 0


So we have n = – 18 or 16


But as n cannot be negative so, we have n = 16


∴ the correct option is (c)


Question 20.

Four numbers are in A.P. and their sum is 72 and the largest of them is twice the smallest. Then the numbers are ………
A. 4, 8, 12, 16

B. 12, 16, 20, 24

C. 10, 12, 14, 16

D. 2, 4, 6, 8


Answer:

We have n = 4


⇒ S4 = 72


Let the 4 nos. Be, a, a + d, a + 2d, a + 3d


Now we have that a + a + d + a + 2d + a + 3d = 72


⇒ 4a + 6d = 72


⇒ 2a + 3d = 36 …….(1)


Now We know that the largest of them is twice the smallest.


So we can say that,


⇒ 2a = a + 3d


⇒ a = 3d


we now put the value of a in (1),


to get :


⇒ 2a + a = 36


⇒ 3a = 36


⇒ a = 12


we know that a is the first term in this AP, so the possible option is (b)


but to check whether it is the correct we will have to check the relation between the first and the last term of the AP of (b) option.


We can see that, in (b), first term is 12


And the last term is 24 so, this is the correct AP.


∴ the correct option is (b)


Question 21.

If S1= 2 + 4 + ... + 2n and S2 = 1 + 3 + ... + (2n — 1), then S1: S2 = …..
A. 

B. 

C. n2

D. (n + 1)


Answer:

We know that, Sn =  × (a + l)


Where a = first term of the AP.


And l = last term of the AP


For S1,


We have S1 =  × (2 + 2n)


⇒ S1 = n × (n + 1)


For S2,


We have S2 =  × (1 + 2n – 1)


⇒ S2 =  × (2n)


⇒ S2 = n2


Now  =  = 


∴  = 


∴ the correct option is (a)


Question 22.

For A.P., Sn — 2Sn – 1 + Sn – 2 =…. (n > 2)
A. 2d

B. d

C. a

D. a + d


Answer:

We can recall that, Sn =  × (2a + (n – 1)d) ……(1)


So now, Sn – 1 =  × (2a + ((n – 1) – 1)d)


⇒ Sn – 1 =  × (2a + (n – 2)d) …….(2)


So now, Sn – 2 =  × (2a + ((n – 2) – 1)d)


⇒ Sn – 2 =  × (2a + (n – 3)d) ………(3)


Now putting the above values in the equation,


We get,


⇒ Sn — 2Sn – 1 + Sn – 2


 × (2a + (n – 1)d) – 2[ × (2a + (n – 2)d)] +  × (2a + (n – 3)d)


 – 2[] + 


= an +  –  – 2an – dn2 + 2a – 2d + 3nd + an +  –  – 2a + 3d


= [an – 2an + an] + [ – dn2 + ] + [2a – 2a] + [ –  + 3nd – ] + [ – 2d + 3d]


= 0 + 0 + 0 + 0 + d


= d


∴ Sn — 2Sn – 1 + Sn – 2 = d


∴ the correct option is (b)


Question 23.

If Sm = n and Sn = m then Sm + n = ……
A. —(m + n)

B. 0

C. m + n

D. 2m — 2n


Answer:

⇒ Sm = n =  × [ 2a + (m – 1)d ]


⇒ = 2a + (m – 1)d ......................(1)


⇒ Sn = m =  × [ 2a + (n – 1)d ]


⇒  = 2a + (n – 1)d ........................(2)


subtracting both equations, we get :


⇒ 2( – ) = d(m – n)


⇒ d = – 2[] ...................(3)


now, Sm + n =  × [ 2a + (m + n – 1)d ]


⇒ Sm + n × = 2a + (m + n – 1)d .......................(4)


now, (4) – (2), we get :


⇒ Sm + n ×  – = d(m)


putting value of d from (3), we get :


⇒ Sm + n ×  –  = – 2[] × m


⇒ Sm + n ×  = – 2[] × m + 


⇒ Sm + n ×  = – 2[] + 


⇒ Sm + n × = – 2 or


⇒ Sm + n = – (m + n)


∴ the correct option is (a).


Question 24.

If T4 = 7 and T7 = 4, then T10 = …..
A. 9

B. 11

C. —11

D. 1


Answer:

Given T4 = 7


And we know that, Tn = a + (n – 1)d


So we have T4 = a + (4 – 1)d


⇒ 7 = a + 3d …… (1)


Then, T7 = 4


⇒ T7 = a + (7 – 1)d


⇒ 4 = a + 6d …….(2)


Now, (2) – (1) gives :


⇒ – 3 = 3d


⇒ d = – 1


so, now, putting value of d in (1),


⇒ 7 = a + 3d


⇒ 7 = a + 3( – 1)


⇒ 7 = a – 3


⇒ a = 10


so now T10 = a + (10 – 1)d


⇒ T10 = a + 9d


= 10 + 9( – 1)


= 10 – 9


= 1


∴ T10 = 1


∴ correct option is (d)


Question 25.

If 2k + 1, 13, 5k — 3 are three consecutive terms of A.P., then k = ……….
A. 17

B. 13

C. 4

D. 9


Answer:

If 2k + 1, 13, 5k — 3 are three consecutive terms of A.P., then,


⇒ 13 – (2k + 1) = (5k – 3 ) – 13 …..(common difference)


⇒ 12 – 2k = 5k – 16


⇒ 28 = 7k


⇒ k = 4


∴ correct option is (c).


Question 26.

(1) + (1 + 1) + (1 + 1 + 1) + ... + (1 + 1 + 1 + ...n — 1 times) =
A. 

B. 

C. n

D. n2


Answer:

(1) + (1 + 1) + (1 + 1 + 1) + .…. + (1 + 1 + 1 + ...n — 1 times)


= 1 + 2 + 3 + 4 ………… + (n – 1)


= Sn – 1


⇒ Sn – 1 =  × [2a + (n – 1 – 1)d]


 × [2(1) + (n – 2)(1)]






∴ the correct option is (a).


Question 27.

In the A.P., 5, 7, 9, 11, 13, 15, ... the sixth term which is prime is
A. 13

B. 19

C. 23

D. 15


Answer:

If we see that, in the given series, d = 2


So if e continue the series, we get :


5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25 …….


We can see that, the prime nos. In the above are :


5, 7, 11, 13, 17, 19, 23, …


so here we see that the sixth term which is prime is 19


∴ the correct option is (b)


Question 28.

For A.P. T18 — T8 = ………
A. d

B. 10d

C. 26d

D. 2d


Answer:

We know that, Tn = a + (n – 1)d


So we have, T18= a + (18 – 1)d


⇒ T18= a + 17d


Now T8= a + (8 – 1)d


⇒ T8= a + 7d


Now we have,


⇒ T18 — T8 = [a + 17d] – [a + 7d]


= 10d


∴ the correct option is (b)


Question 29.

If for A.P., T25 — T20 = 15 then d =
A. 3

B. 5

C. 20

D. 25


Answer:

We know that, Tn = a + (n – 1)d


So we have, T25= a + (25 – 1)d


⇒ T25= a + 24d


Now we have, T20= a + (20 – 1)d


⇒ T20= a + 19d


It is given that, T25 — T20 = 15


So we can say that,


⇒ 15 = [a + 24d] – [a + 19d]


⇒ 15 = 5d


⇒ d = 3


∴ the correct option is (a).


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