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Class 11 RD Sharma Solutions – Chapter 19 Arithmetic Progressions- Exercise 19.4 | Set 3
  • Last Updated : 18 Mar, 2021

Question 23. If the 5th and 12th terms of an A.P. are 30 and 65 respectively, what is the sum of the first 20 terms?  

Solution:

According to question we have

a5 = 30 

Using the formula

⇒ a + (5 – 1)d = 30 



⇒ a + 4d = 30        ……………. (i) 

Also, an = 65 

Using the formula

⇒ a + (12 – 1)d = 65 

⇒ a + 11d = 65      ……………(ii) 

On solving eq(i) and (ii), we get 

7d = 35 

⇒ d = 5 

On putting the value of d in (i), we get : 

a + 4 x 5 = 30 

⇒ a = 10 

Now we find the sum of first 20 terms

S20 = 20/2 [2 x 10 + (20 – 1) x 5] 

⇒ S20 = 10 [20 + 95]

⇒ S20 = 1150

Question 24. Find the sum of n terms of the A.P. whose kth terms is 5k + 1.  

Solution:

According to question we have

ak = 5k + 1 



For k = 1, a1 = 5 x 1 + 1 = 6 

For k = 2, a2 = 5 x 2 + 1 = 11 

For k = n, an = 5n + 1 

Now we find the sum of n terms

Sn = n/2 [a + an] 

⇒ Sn = n/2[6 + 5n + 1] = n/2 (5n + 7)  

Question 25. Find the sum of all two-digit numbers which when divided by 4, yields 1 as remainder. 

Solution:

According to question we have

A.P = 13, 17….97

From the A.P we have

a = 13, d = 4, an = 97 

Now using the formula

⇒ an = a (n – 1)d 

⇒ 97 = 13 + (n – 1)4 

⇒ 84 = 4n – 4 

⇒ 88 = 4n 

⇒ 22 = n          …………… (1) 

Now we find the sum of all two-digit numbers using the given formula

Sn = n/2[2a + (n – 1)d] 

S22 = 22/2 [2 x 13 + (22 – 1) x 4]      ( From eq(1)) 



⇒ S22 = 11[26 + 84]

⇒ S22 = 11[110] = 1210

Question 26. If the sum of a certain number of terms of the AP 25, 22, 19, … is 116. Find the last term. 

Solution:

According to question 

A.P. = 25, 22, 19 

From the A.P we have

a = 25, d = 22 – 25 = -3 

Sn = 116 

Now using the formula

⇒ n/2 [2a + (n – 1)d] = 116 

⇒ n [2 x 25 + (n – 1)(-3)] = 232 

⇒ 50n -3n2 + 3n = 232 

⇒ 3n2 – 53n + 232 = 0 

⇒ 3n2 – 29n – 24n + 232 = 0 

⇒ n(3n – 29) – 8(3n – 29) = 0 

⇒(3n – 29)(n – 8) = 0 

⇒ n = 29/ 3 or 8 

Since n cannot be a fraction, n = 8. 

Now we find the last term using the following formula

an = a + (n – 1)d

⇒ a8 = 25 + (8 – 1)(-3)

⇒ a8 = 4

Question 27. Find the sum of odd integers from 1 to 2001.  

Solution:

According to question 

A.P = 1, 3, 5 … 2001

From the A.P we have

a = 1 and d = 2 

an = 2001 

Now using the formula

⇒ 1 + (n – 1)2 = 2001 

⇒ 2n – 2 = 2000 

⇒ 2n = 2002 

⇒ n = 1001  

Now we find the sum of odd integers from 1 to 2001

Also, S1001= 1001/2 [2 x 1 + (1001 – 1)2]

⇒ S1001 = 1001/2 [2 + 2000]

⇒ S1001 = 1001 x 1001 = 1002001 

Question 28. How many terms of the A.P. -6, -11/2, -5, … are needed to give the sum -25? 

Solution:

According to question 

A.P = -6, -11/2, -5, … 

From the A.P we have

a = – 6 and d =-11/2 – (-6) = 1/2

Sn = -25  

Now using the formula

 ⇒ -25 = n/2 [2 x (-6) + (n -1)(1/2)]

⇒ -25 = n/2 [ -12 + n/2 – 1/2]

⇒ -50 = n [ n/2 – 25/2]

⇒ -100 = n [ n – 25]

⇒ n2 – 25n + 100 = 0

⇒ (n – 20)(n – 5) = 0

⇒ n = 20 or n = 5

Question 29. In an A.P. the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is -112.  

Solution:

According to question we have

a = 2, S5 = 1/4(S10 – S5)  

S5 =  5/2 [2 x 2 + (5 – 1)d]  

⇒ S5 = 5 [2 + 2d]    ………………..(i) 

Also, S10 = 10/2 [2 x 2 + (10 – 1)d] 

⇒ S10 = 5[4 + 9d]   ………………..(ii) 

Since S5 = 1/4 (S10 – S5

So, from eq(i) and (ii), we have: 

⇒ 5[2 + 2d] = 1/4 [5(4 + 9d) – 5(2 + 2d)] 

⇒ 8 + 8d = 4 + 9d – 2 – 2d  

⇒ d = -6

Now we find the 20th term

a20 = a + (20 – 1)d 

⇒ a20 = a + 19d

⇒ a20 = 2 + 19(-6) = -112

Hence proved

Question 30. If S1 be the sum of (2n + 1) terms of an A.P. and S2 be the sum of its odd terms, the prove that: S1 : S2 = (2n + 1) : (n + 1) 

Solution:

Let us assume A. P. be a, a + d, a + 2d… 

So, S1 = (2n + 1)/2 [2a + (2n + 1 – 1)d] 

Now using the formula, we get

⇒ S1 = (2n + 1)/2 [2a + (2n)d] 

⇒ S1 = (2n + 1)(a + nd)          …………………(i) 

Now using the formula, we get

S2 = (n + 1)/2 [2a + (n + 1 – 1) x 2d] 

⇒ S2 = (n +1)/2 [2a + 2nd] 

⇒ S2 = (n + 1)[a + nd]           …………………..(ii)

 From eq(i) and (ii), we get : 

S1 / S2 = (2n + 1) / (n + 1) 

Hence, proved 

Question 31. Find an A.P. in which the sum of any number of terms is always three times the squared number of these terms. 

Solution:

Given that Sn = 3n2

So, for n = 1, S1 = 3 x 12 = 3 

For n = 2, S2 = 3 x 22 = 12 

For n = 3, S3= 3 x 32 = 27 and so on  

So, S1 = a1 = 3 

a2 = S2 – S1 = 12 – 3 = 9 

a3 = S3 – S2 = 27 – 12 = 15 and so on  

Hence, the A.P. = 3, 9, 15… 

Question 32. If the sum of n terms of an A.P. is nP + 1/2 – n (n – 1) Q, where P and Q are constants, find the common difference.

Solution:

According to question we have

Sn = nP + 1/2 n(n – 1)Q 

For n = 1, S1 = P + 0 = P 

For n = 2, S2 = 2P + Q 

Also, a1 = S1 = P,

a2 = S2    S1 = 2P + Q – P = P + Q 

Hence, the common difference d = a2 – a1 = P + Q – P = Q  

Question 33. The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18th terms.

Solution:

Let us considered we have two A.P’s. So, a1 and a2 are the first terms and d1 and d2 is common difference of the A.P’s

According to question we have

(5n + 4) / (9n + 6) = (Sum of n terms in the first A.P.) / (Sum of n terms in the second A.P.) 

⇒ (5n + 4) / (9n + 6) = (2a1+ [(n — 1)d1]) / (2a2 + [(n — 1)d2]  ………(1)

Now put n = 2 x 18 – 1 = 35 in eq(1), we get 

(5 x 35 + 4) / (9 x 35 + 6) = (2a1 + 34d1) / (2a2 + 34d2)

179 /321 = (a1 + 17d1) / (a2 + 17d2) = (18th term of the first A.P.) / (18th term of the second A.P.)

Hence, the ratio of 18th terms = (a1 + 17d1) / (a2 + 17d2) = 179 /321

Question 34. The sums of n terms of two arithmetic progressions are in the ratio 7n + 2 : n + 4. Find the ratio of their 5th terms.

Solution:

Let us considered we have two A.P’s. So, a1 and a2 are the first terms and S1 and S2 are the sum of the first n terms. 

According to question we have

S1 = n/2 [2a1 + (n – 1)d1

And, S2 = n/2 [2a2 + (n -1)d2

Given:  

S1 / S2 = (n/2 [2a1 + (n – 1)d1])/(n/2 [2a2 + (n – 1)d2]) = (7n + 2) / (n + 4)

Now we find the ratio of their 5th terms

[2a1 + (9 – 1)d1] / [2a2 + (9 – 1)d2] = (7(9) + 2) / (9 + 4)

[2a1 + (8)d1] / [2a2 + (8)d2] = 65/13

[a1 + (4)d1] / [a2 + (4)d2] = 5/1 = 5 : 1

Hence, the ratio of 5th terms = 5 : 1

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