Question

If t10 = 41 a = 5, find d and tn.​

Answer 1

Given:

$t_{10}$ = 41 and a = 5

We have to find, the values of d and $t_n$.​

Solution:

We know that,

The nth term of an AP.,

$t_n$ = a + (n - 1)d

∴ $t_{10}$ = 5 + (10 - 1)d = 41

⇒ 9d = 41 - 5 = 36

⇒ d = 4

∴ The nth term of an AP.,

$t_n$ = 5 + (n - 1)4

⇒ $t_n$ = 5 + 4n - 4

⇒ $t_n$ = 4n + 1

Thus, d = 4 and $t_n$ = 4n + 1