Question

Find the sum (5 + 13 + 21 + ... + 181).

Answer 1

Here, a = 5, d = (13 - 5) = 8 and l = 181
Let the total number of terms be n.
Then T_n = 181
⇒ a + (n - 1)d = 181
⇒ 5 + (n - 1​) ⨯​ 8 = 181
⇒ 8n = 184
⇒ n = 23
Hence, there are 23 terms in the AP.
∴ Required sum = $\frac{n}{2}\left(a+l\right)$

= $\frac{23}{2}\left(5+181\right)=23\times 93=2139$
Hence, the required sum is 2139.