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

${\displaystyle \Rightarrow }$ a + 39d = a + 15d + 72

${\displaystyle \Rightarrow }$ 39d – 15d = 72

${\displaystyle \Rightarrow }$ 24d = 72

${\displaystyle \Rightarrow }$ d = ${\displaystyle \frac{72}{24}}$ = 3