# Class 11 RD Sharma Solutions – Chapter 19 Arithmetic Progressions- Exercise 19.4 | Set 2

Last Updated : 30 Aug, 2022

### Question 11. Find the sum of all integers between 100 and 550 which are divisible by 9.

Solution:

According to question

A.P = 108, 117…549

From the given A.P. we get

a(first term) = 108, d(common difference) = 9, an(nth term) = 549

Find the value of n using the given formula

an = a + (n – 1)d

549 = 108 + (n – 1)(9)

441 = 9n – 9

450 = 9n

n = 50

Now, we find the sum of the given A.P. using the following formula

S = n/2 [2a + (n -1)d]

S = 50/2 [2(108) + (50 – 1)(9)]

= 25 [216 + (49)(9)]

= 25 [216 + 441]

= 25 [657]

= 16425

Hence,  the sum of all integers between 100 and 550 which are divisible by 9 = 16425

### 3 + 5 + 7 + 6 + 9 + 12 + 9 + 13 + 17 + … to 3n terms.

Solution:

A.P = 3 +5 +7 + 9 + … to 3n

From the given A.P. we get

a = 3, d = 2, n = 3n

Now, we find the sum of the given A.P. using the following formula

S = n/2 [2a + (n – 1)d]

= 3n/2 [2(3) + (3n – 1)(2)]

= 3n [3 + (3n – 1)]

= 3n [3n + 2]

Hence, the sum of the given A.P = 3n [3n + 2]

### Question 13. Find the sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7.

Solution:

According to question

A.P = 103, 119,…,791

From the given A.P. we get

a = 103, l = 791

Find the value of n using the given formula

an = a + (n – 1)d

791 = 103 + (n – 1)16

n = 44

Now, we find the sum of the given A.P. using the following formula

S = n/2 [a + l]

S = 44/2 [103 + 791]

= 22 [894]

= 19668

Hence, the sum of all those integers between 100 and 800

each of which on division by 16 leaves the remainder 7 = 19668

### (ii) 1 +4+7+ 10 + … + x = 590

Solution:

(i) A.p= 25 + 22 + 19 + 16 +…+x = 115

From the given A.P. we get

a = 25, d = -3, S = 115

Using the formula

S = n/2[2a + (n – 1)d]

â‡’ 115 = n/2 [2 x 25+ (n – 1)(-3)]

â‡’ 115 x 2 = n[50 – 3n + 3]

â‡’ 230 = n(53 – 3n)

â‡’ 230 = 53n – 3n2

â‡’ 3n2 – 53n + 230 = 0

Now, Put the value of a = 3, b = – 53 and c = 230, we get

n = 46/6,10

â‡’ n = 10 as n â‰  46/6

So, an = x = a + (n – 1)d

â‡’ x = 25 + (10 – 1)(-3)

â‡’ x = 25 – 27 = -2

Hence, the value of x = -2

(ii) A.P = 1 + 4 + 7 + 10 +….+ x = 590

From the given A.P. we get

a = 1, d = 3

Using the formula

S = n/2 [2a + (n – 1)d]

â‡’ 590 = n/2[2 x 1 + (n – 1)(3)]

â‡’ 590 x 2 = n[2 + 3n – 3]

â‡’ 1180 = n(3n – 1)

â‡’ 1180 = 3n2 – n

â‡’ 3n2 – n -1180 = 0

Now, Put the value of a = 3, b = – 1 and c = -1180, we get

n = -118/6, 20

â‡’ n = 20, as n â‰  -118/6

an = x = a + (n – 1)d

â‡’ x = 1 + (20 – 1)(3)

â‡’ x = 1 + 60 – 3 = 58

Hence, the value of x = 58

### Question 15. Find the rth term of an A.P., the sum of whose first n terms is 3n2 + 2n.

Solution:

According to the question

As, Sn = 3n2 + 2n

So, a = S1 = 3 x 12 + 2 x 1 = 3 + 2 = 5 and

S2 = 3 x 22+2 x 2 = 12+4 = 16

â‡’ a + a2 = 16

â‡’ a + a + d = 16

â‡’ 2a + d = 16

â‡’ 2 x 5 + d = 16

â‡’ d = 16 – 10

â‡’ d = 6

Now,

ar = a + (r – 1)d

= 5+ (r -1) x 6

= 5 + 6r – 6

Hence, the ar = 6r – 1

### Question 16. How many terms are there in the A.P. whose first and fifth terms are -14 and 2 respectively and the sum of the terms is 40?

Solution:

According to question we have

a = -14 and Sn = 40 ………………………. (i)

a5 = 2

By using formula

â‡’ a + (5 – 1)d = 2

â‡’ -14 + 4d = 2

â‡’ 4d = 16

â‡’ d = 4  …………………………..(ii)

By using formula

Sn = n/2 [2a + (n – 1)d]

â‡’ 40 = n/2 [2(-14) + (n – 1) x 4]                   ( From eq(i) and (ii))

â‡’ 80 = n[-28 + 4n – 4]

â‡’ 80 = 4n2 – 32n

â‡’ n2 – 8n – 20 = 0

â‡’ (n -10)(n + 2) = 0

â‡’ n = 10, -2

But n cannot be negative.

Hence, the total number of terms in the A.P = 10

### Question 17. The sum of the first 7 terms of an A.P. is 10 and that of the next 7 terms is 17. Find the progression.

Solution:

According to question we have

S7 = 10

By using formula

Sn = n/2 [2a + (n – 1)d]

â‡’ 7/2 [2a + (7 – 1)d] = 10

â‡’ 7/2 [2a + 6d] = 10

â‡’ a + 3d = 10/7   ………………………………… (i)

Also, the sum of the next seven terms = S14 – S7 = 17

â‡’ 14/2 [2a + (14 – 1)d] – 7/2 [2a + (7 – 1)d] = 17

â‡’ 7[2a + 13d] – 7/2[2a + 6d] = 17

14a + 91d – 7a – 21d = 17

7a + 70d = 17

a + 10d = 17/7  ………………………………….. (ii)

From eq(i) and (ii), we get:

10/7 – 3d = 17/7 – 10d

â‡’ 7d = 1

â‡’ d = 1/7

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

a + 3d = 10/7

â‡’ a+ 3/7 = 10/7

â‡’ a = 1, d = 1/7

Hence, the progression = 1, 8/7, 9/7, 10/7 …

### Question 18. The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference, and the sum of first 20 terms.

Solution:

According to question we have

a3 = 7, a7 – 3a3 = 2

By using formula

an = a + (n – 1)d

â‡’ a + (3 – 1)d = 7

â‡’ a + 2d = 7  …………………………………… (i)

Also,

a7 – 3a3 = 2

â‡’ a7 – 21 = 2(Given)

â‡’ a + (7 – 1)d = 23

â‡’ a + 6d = 23 ………………………………………. (ii)

From eq(i) and (ii), we get

4d = 16

â‡’ d = 4

On putting the value in eq(i), we get

a + 2(4) = 7

a = -1

By using formula

Sn = n/2 [2a + (n – 1)d]

S20 = 20/2 [2(-1) + (20-1)(4)]

â‡’ S20 = 10[-2 + 76]

â‡’ S20 = 10[74] = 740

Hence,

a = -1, d = 4, S20 = 740

### Question 19. The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference.

Solution:

According to question we have

a = 2,l = 50, Sn = 442

Now, Sn = 442

â‡’ n/2[a + l] = 442

â‡’ n/2 [2 + 50] = 442

â‡’ n = 17

a17 = 50

By using formula, we get

â‡’ a + (17 – 1) d = 50

â‡’ 2 + 16d = 50

â‡’ d = 3

### Question 20. The number of terms of an A.P. is even; the sum of odd terms is 24, of the even terms is 30, and the last term exceeds the first by , find the number of terms and the series.

Solution:

According to question, we have

a1 + a3+… +a2n-1 = 24 ……………..(1)

a2 + a4+… +a2n = 30 ……………..(2)

Now, subtracting eq(1) from (2), we get:

(d + d+…+ upto n terms) = 6

â‡’ nd = 6 ………….(3)

Given :

a2n = a1 + 21/2

â‡’ a2n – a1 = 21/2

â‡’ a + (2n – 1)d – a = 21/2  [a2n= a + (2n – 1)d, a1 = a]

â‡’ 2nd – d = 21/2

â‡’ 2 x 6 – d = 21/2 (From eq(3))

â‡’ d = 3/2

On putting the value in eq(3), we get

n = 4

â‡’ 2n = 8

Thus, there are 8 terms in the progression.

To find the value of the first term:

a2 + a4+…+a2n = 30

â‡’ (a + d) + (a + 3d)+… +[a + (2n – 1)d] = 30

â‡’ n/2 [(a + d) + a + (2n – 1)d] = 30

On putting n = 4 and d = 3/2, we get

a = 3/2

Hence, the series will be 1, 1/2, 3,  …

### Question 21. If Sn = n2p and Sm = m2 p, m â‰  n, in an A.P., prove that Sp = p3.

Solution:

Sn = n2p

â‡’ n/2 [2a + (n – 1)d] = n2p

â‡’ 2np = 2a + (n – 1)d          ………………. (i)

Sm = m2p

â‡’ n/2 [2a + (m – 1)d] = m2p

â‡’ 2mp = 2a + (m – 1)d       ……………….. (ii)

On subtracting eq(ii) from eq(i), we get

2p(n – m) = (n – m)d

2p = d                                ………………..(iii)

On substituting the value in eq(i), we get

nd = 2a + (n – 1)d

â‡’ nd – nd + d = 2a

â‡’ a = d/2 = p [from eq(iii)]  ……………….(iv)

Sp = p/2[2a + (p – 1)d]

â‡’ Sp = p/2[2p + (p – 1)2p]

â‡’ Sp = p/2[2p + 2p2 -2p]

â‡’ Sp = p/2[2p2]

â‡’ Sp = p3

Hence Proved

### Question 22. If 12th term of an A.P. is -13 and the sum of the first four terms is 24, what is the sum of first 10 terms?

Solution:

Let us considered first term = a and common difference = d

Given that a12 = -13

Using the formula

â‡’ a + (12 – 1)d = -13

â‡’ a + 11d = -13           ………………. (i)

Also, S4 = 24 (given)

Using the formula

â‡’ 4/2 [2a + (4 – 1)d] = 24

â‡’ 2(2a + 3d) = 24

â‡’ 2a + 3d = 12           ………………. (ii)

From eq(i) and (ii), we get

19d = -38

d = -2

Now put the value of d in eq(i), we get

a + 11(-2) = -13

â‡’ a = 9

S10 = 10/2 [(2)(9) + (10 – 1)(-2)]

â‡’ S10 = 5[18 – 18] = 0

Hence, the sum of first ten terms = 0

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