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What is the sum of first 24 terms of the sequence whose $n^{th}$ term is given by

$a_{n} = 3 + \frac{2 n}{3}$ ?

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278

272

270

268

answer is B.

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Detailed Solution

We are given that,

a_{n} = 3 + \frac{2 n}{3}

.
So, we get

a_{n + 1} = 3 + \frac{2 n + 2}{3}

.
Therefore, the common difference,

d = a_{n + 1} - a_{n}

\Rightarrow d = 3 + \frac{2 n + 2}{3} - 3 - \frac{2 n}{3}

\Rightarrow d = \frac{2}{3}

First term,

a = a_{1}

\Rightarrow a = 3 + \frac{2}{3}

\Rightarrow a = \frac{11}{3}

We know that, sum of

n

first terms of an arithmetic progression is given as

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

On substituting the values, we get

\Rightarrow S_{24} = \frac{24}{2} [\frac{22}{3} + 23 (\frac{2}{3})]

\Rightarrow S_{24} = 12 [\frac{22 + 46}{3}]

\Rightarrow S_{24} = \frac{12 \times 68}{3}

\Rightarrow S_{24} = 4 \times 68

\Rightarrow S_{24} = 272

Hence, the sum of first

24

terms of the given series is

272

.
Hence, the correct option is 2.

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