# Class 11 RD Sharma Solutions – Chapter 19 Arithmetic Progressions- Exercise 19.4 | Set 3

### 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

a

_{5}= 30Using the formula

â‡’ a + (5 – 1)d = 30

â‡’ a + 4d = 30 ……………. (i)

Also, a

_{n}= 65Using the formula

â‡’ a + (12 – 1)d = 65

â‡’ a + 11d = 65 ……………(ii)

On solving eq(i) and (ii), we get

7d = 35

â‡’ d = 5

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

a + 4 x 5 = 30

â‡’ a = 10

Now we find the sum of first 20 terms

S

_{20}= 20/2 [2 x 10 + (20 – 1) x 5]â‡’ S

_{20 }= 10 [20 + 95]â‡’ S

_{20}= 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

a

_{k}= 5k + 1For k = 1, a

_{1}= 5 x 1 + 1 = 6For k = 2, a

_{2}= 5 x 2 + 1 = 11For k = n, a

_{n}= 5n + 1Now we find the sum of n terms

Sn = n/2 [a + an]

â‡’ 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, a

_{n}= 97Now using the formula

â‡’ a

_{n}= a (n – 1)dâ‡’ 97 = 13 + (n – 1)4

â‡’ 84 = 4n – 4

â‡’ 88 = 4n

â‡’ 22 = n …………… (1)

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

S

_{n}= n/2[2a + (n – 1)d]S

_{22}= 22/2 [2 x 13 + (22 – 1) x 4] ( From eq(1))â‡’ S

_{22 }= 11[26 + 84]â‡’ S

_{22 }= 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

S

_{n}= 116Now using the formula

â‡’ n/2 [2a + (n – 1)d] = 116

â‡’ n [2 x 25 + (n – 1)(-3)] = 232

â‡’ 50n -3n

^{2}+ 3n = 232â‡’ 3n

^{2}– 53n + 232 = 0â‡’ 3n

^{2}– 29n – 24n + 232 = 0â‡’ n(3n – 29) – 8(3n – 29) = 0

â‡’(3n – 29)(n – 8) = 0

â‡’ n = 29/ 3 or 8

Since n cannot be a fraction, n = 8.

Now we find the last term using the following formula

a

_{n }= a + (n – 1)dâ‡’ a

_{8 }= 25 + (8 – 1)(-3)â‡’ a

_{8}= 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

a

_{n}= 2001Now using the formula

â‡’ 1 + (n – 1)2 = 2001

â‡’ 2n – 2 = 2000

â‡’ 2n = 2002

â‡’ n = 1001

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

Also, S

_{1001}= 1001/2 [2 x 1 + (1001 – 1)2]â‡’ S

_{1001}= 1001/2 [2 + 2000]â‡’ S

_{1001 }= 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

S

_{n}= -25Now using the formula

â‡’ -25 = n/2 [2 x (-6) + (n -1)(1/2)]

â‡’ -25 = n/2 [ -12 + n/2 – 1/2]

â‡’ -50 = n [ n/2 – 25/2]

â‡’ -100 = n [ n – 25]

â‡’ n

^{2 }– 25n + 100 = 0â‡’ (n – 20)(n – 5) = 0

â‡’ 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, S

_{5 }= 1/4(S_{10}– S_{5})S

_{5 }= 5/2 [2 x 2 + (5 – 1)d]â‡’ S

_{5}= 5 [2 + 2d] ………………..(i)Also, S

_{10}= 10/2 [2 x 2 + (10 – 1)d]â‡’ S

_{10}= 5[4 + 9d] ………………..(ii)Since S

_{5}= 1/4 (S_{10}– S_{5})So, from eq(i) and (ii), we have:

â‡’ 5[2 + 2d] = 1/4 [5(4 + 9d) – 5(2 + 2d)]

â‡’ 8 + 8d = 4 + 9d – 2 – 2d

â‡’ d = -6

Now we find the 20th term

a

_{20 }= a + (20 – 1)dâ‡’ a

_{20}= a + 19dâ‡’ a

_{20 }= 2 + 19(-6) = -112Hence proved

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

**Solution:**

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

So, S

_{1}= (2n + 1)/2 [2a + (2n + 1 – 1)d]Now using the formula, we get

â‡’ S

_{1}= (2n + 1)/2 [2a + (2n)d]â‡’ S

_{1}= (2n + 1)(a + nd) …………………(i)Now using the formula, we get

S

_{2}= (n + 1)/2 [2a + (n + 1 – 1) x 2d]â‡’ S

_{2 }= (n +1)/2 [2a + 2nd]â‡’ S

_{2}= (n + 1)[a + nd] …………………..(ii)From eq(i) and (ii), we get :

S

_{1}/ S_{2}= (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 S

_{n }= 3n^{2}So, for n = 1, S

_{1}= 3 x 1^{2}= 3For n = 2, S

_{2}= 3 x 2^{2}= 12For n = 3, S

_{3}= 3 x 3^{2 }= 27 and so onSo, S

_{1 }= a_{1 }= 3a

_{2}= S_{2}– S_{1}= 12 – 3 = 9a3 = S

_{3}– S_{2}= 27 – 12 = 15 and so onHence, 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

S

_{n }= nP + 1/2 n(n – 1)QFor n = 1, S

_{1 }= P + 0 = PFor n = 2, S

_{2}= 2P + QAlso, a

_{1 }= S_{1 }= P,a

_{2}= S_{2 }–_{ }S_{1}= 2P + Q – P = P + QHence, the common difference d = a

_{2}– a_{1}= 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, a

_{1}and a_{2}are the first terms and d_{1}and d_{2}is common difference of the A.P’sAccording 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.)

â‡’ (5n + 4) / (9n + 6) = (2a

_{1}+ [(n â€” 1)d_{1}]) / (2a_{2}+ [(n â€” 1)d_{2}] ………(1)Now put n = 2 x 18 – 1 = 35 in eq(1), we get

(5 x 35 + 4) / (9 x 35 + 6) = (2a

_{1}+ 34d_{1}) / (2a_{2}+ 34d_{2})179 /321 = (a

_{1 }+ 17d_{1}) / (a_{2}+ 17d_{2}) = (18th term of the first A.P.) / (18th term of the second A.P.)Hence, the ratio of 18th terms = (a

_{1 }+ 17d_{1}) / (a_{2}+ 17d_{2}) = 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, a

_{1}and a_{2}are the first terms and S_{1}and S_{2}are the sum of the first n terms.According to question we have

S

_{1}= n/2 [2a_{1}+ (n – 1)d_{1}]And, S

_{2}= n/2 [2a_{2}+ (n -1)d_{2}]Given:

S

_{1 }/ S_{2 }= (n/2 [2a_{1}+ (n – 1)d_{1}])/(n/2 [2a_{2}+ (n – 1)d_{2}]) = (7n + 2) / (n + 4)Now we find the ratio of their 5th terms

[2a

_{1}+ (9 – 1)d_{1}] / [2a_{2}+ (9 – 1)d_{2}] = (7(9) + 2) / (9 + 4)[2a

_{1}+ (8)d_{1}] / [2a_{2}+ (8)d_{2}] = 65/13[a

_{1}+ (4)d_{1}] / [a_{2}+ (4)d_{2}] = 5/1 = 5 : 1Hence, the ratio of 5th terms = 5 : 1

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