1. EX. NO. 4

2. 1. Find the sum of the first n natural numbers and
hence find the sum of first 20 natural numbers.
Sol. The first n natural numbers are 1, 2, 3, 4, ......n
These natural number form an A.P. with a = 1, d = 1
t1 = 1 and tn = n
Alternative method :
Sn = n/2 [t1 + tn]
S20 = 20/2[t1 + t20]
= 10 [1 + 20]
= 10 [21]
S20 = 210
The sum of first twenty terms is 210.

3. 2. Find the sum of all odd natural numbers
from 1 to 150.
Sol. The odd natural numbers from 1 to 150
are 1, 3, 5, 7, 9, .........., 149.
These numbers form an A.P. with a = 1, d = 2
Let, 149 be nth term of an A.P.
tn = a + (n – 1) d
149 = 1 + (n – 1) 2 = 1 + 2n – 2 = 2n – 1
149 + 1 = 2n
2n = 150
n = 75
∴ 149 is 75th term of A.P.

4. Sn =
n/
2
[t1 + tn]
S75 = 75/2[t1 + t75]
=
75/
2[
[1 + 149]
= 75/2[ [150]
S75 = 75 x 75 =5625

5. 3. Find S10 if a = 6 and d = 3.
Sol. For an A.P. a = 6, d = 3
n/ [2a + (n – 1)d]
Sn = 2
S10 = 10/2 [2(6) + (10 – 1 ) 3]
S10 = 5 [2 (6) + 9 (3)]
S10 = 5 (12 + 27)
S10 = 5 (39)
S10 = 195

6. 4. Find the sum of all numbers from 1 to 140
which are divisible by 4.
Sol. The natural numbers from 1 to 140 that
are divisible by 4 are 4, 8, 12, 16, .............., 140
Here, a = 4, d = t2 – t1 = 8 – 4 = 4 and tn =
140
tn = a + (n – 1) d
∴ 140 = 4 + (n – 1) 4
∴ 140 = 4 + 4n – 4
∴ 140 = 4n
∴ n = 140 /4
∴ n = 35

7. n/ [2a + (n – 1)d]
Sn = 2
35/ [2a + (35 – 1 ) d]
∴ S35 = 2
35/ [2 (4) + 34 (4)]
∴ S35 = 2
35/ (8 + 136)
∴ S35 = 2
35/ (144)
∴ S35 = 2
∴ S35 = 35 x 72
∴ S35 = 2520

8. 5. Find the sum of the first n odd natural
numbers. Hence find 1 + 3 + 5 + ... + 101.
Sol. The first n odd natural numbers are 1,
3, 5, 7, ............., n
Here, a = 1, d = t2 – t1 = 3 – 1 = 2
Sn = n/2 [2a + (n – 1)d]
∴ Sn = n/2 [2(1) + (n – 1)(2)]
∴ Sn = n/2 (2 + 2n – 2)
∴ Sn = n/2 (2n)
∴ Sn = n2 ........ eq. (1)

9. Now, we have
1 + 3 + 5 + ........ + 101
Let, tn = 101
tn = a + (n – 1) d
∴ 101 = a + (n – 1) d
∴ 101 = 1 + (n – 1) 2
∴ 101 = 1 + 2n – 2
∴ 101 = 2n – 1
∴ 101 + 1 = 2n
∴ 2n = 102
∴ n = 102/2
∴ n = 51

10. Now, 101 is the 51st term of A.P.,
We have to find sum of 51 terms i.e.
S51,
2 [From (i)]
Sn = n
∴ S51 = (51)2
∴ S51 = 2601

11. 6. Obtain the sum of the 56 terms of an
A. P. whose 19th and 38th terms are
52 and 148 respectively.
Sol. t19 = 52, t38 = 148
tn = a + (n – 1) d
∴ t19 = a + (19 – 1) d
∴ 52 = a + 18d ......(i)
∴ t38 = a + (38 – 1) d
∴ 148 = a + 37d ......(ii)

12. Adding (i) and (ii) we get,
a + 18d = 52
a + 37d = 148 .
2a + 55d = 200 . ..........Eq. no. (iii)
Sn = n/2[2a + (n – 1)d]
∴ S56 = 56/2 [2a + (56 – 1) d]
∴ S56 = 28[2a + 55d]
∴ S56 = 28[200] [from eq. no. (iii)
∴ S56 = 5600
∴ Sum of first 56 terms of A.P. is 5600.

13. 7. The sum of the first 55 terms of an
A. P. is 3300. Find the 28th term.
Sol. S55 = 3300 [Given]
Sn = n/2[2a + (n – 1)d]
∴ S55 = 55/2[2a + (55 – 1) d]
∴ 3300 = 55/2[2a + 54d]
55/ × (2)[a + 27d]
∴ 3300 = 2
∴ 3300 = 55[a + 27d]
3300/ = a + 27d
55
∴ a + 27d= 60 ......eq. no. (1)

14. But,
tn = a + (n – 1) d
∴ t28 = a + (28 – 1) d
∴ t28 = a + 27d
∴ t28 = 60 [From (i)]
∴ Twenty eighth term of A.P.
is 60.

15. 8. Find the sum of the first n even natural
numbers. Hence find the sum of first 20 even
natural numbers.
Sol. The first n even natural numbers are 2, 4, 6,
8, .....
Here, a = 2, d = t2 – t1 = 4 – 2 = 2
∴ Sn = n/2[2a + (n – 1)d]
∴ Sn = n/2[2 (2) + (n – 1) 2]
∴ Sn = n/2[4 + 2n – 2]
∴ Sn = n/2[2n + 2]
∴ Sn =n/2 ×(2) (n + 1)
∴ Sn = n (n + 1)

16. ∴ S20 = 20 (20 + 1)
∴ S20 = 20 (21)
∴ S20 = 420
∴ Sum of first twenty
even natural numbers is
420