(5 + 13 + 21 + ... + 181) = ?
(a) 2476
(b) 2337
(c) 2219
(d) 2139

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Solution

(d) 2139

Here, a = 5, d = (13-5) = 8 and l = 181
Let the number of terms be n.
Then T_n = 181
⇒a + (n-1) d = 181
⇒ 5 + ( n -1) ⨯ 8 = 181
⇒ 8n = 184
⇒ n = 23

∴ Required sum = $\frac{n}{2} (a + l)$
$= \frac{23}{2} (5 + 181) = 23 \times 93 = 2139$
Hence, the required sum is 2139.

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