Question

Solve:

1 + 6 + 11 + 16 + ⋯ + x = 148

Answer 1

Clearly, terms of the given series form an A.P. with first term a = 1 and common difference d = 5.

Let there be n terms in this series. Then,

1 + 6 + 11 +16 +… + x = 148

⇒ sum of n terms = 148

⇒ \(\frac{n}{2}\)[2a + (n − 1)d)] = 148

⇒ \(\frac{n}{2}\)[2 + (n − 1)5)] = 148

⇒ 5n² − 3n − 296 = 0

⇒ (n − 8)(5n + 37) = 0

⇒ n = 8       [∴ n is not negative]

Now, x = n^th term

⇒ x = a + (n − 1)d

⇒ x = a + (8 − 1) × 5 = 36

[∴ a = 1, d = 5, n = 8]