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 as follows: 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 ......(i)
Now, we have,
1 + 3 + 5 + ........ + 101
Let, 101 be the nth term of A.P.
∴ 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 = 51
∴ 101 is the 51st term of A.P.,
Now, We have to find sum of 51 terms i.e. S51,
Sn = n2 [From (i)]
∴ S51 = (51)2
∴ S51 = 2601
