Question

The 40th term of an A.P. exceeds its 16th term by 72. Then its common difference is ______.

पर्याय

• 40

• 40 – 16

• 72

• 3

Answer 1

The 40th term of an A.P. exceeds its 16th term by 72. Then its common difference is 3.

Explanation:

Let a be the first term and d be the common difference of given A.P.

Given T₄₀ = T₁₆ + 72

`\implies` a + 39d = a + 15d + 72

`\implies` 39d – 15d = 72

`\implies` 24d = 72

`\implies` d = `72/24` = 3