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The sum of $12$ terms of an AP, whose first term is $4,$ is $256 .$ What is the last term?

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35

36

37

None

answer is D.

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Detailed Solution

In the given AP,

n = 12, a = 4

and

S_{12} = 256

.
As we know, sum of

n

terms of an AP is given by,

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

\Rightarrow 256 = \frac{12}{2} [2 \times 4 + (12 - 1) d]

\Rightarrow 256 = 6 [8 + 11 d]

\Rightarrow 256 = 48 + 66 d

\Rightarrow 208 = 66 d

\Rightarrow d = \frac{104}{33}

Also, the

n^{th}

term of an AP is given by,

t_{n} = a + (n - 1) d

⇒

t_{12} = 4 + (12 - 1) \frac{104}{33}

= 4 + 11 \times \frac{104}{33}

= \frac{12 + 104}{3}

⇒

t_{12} = \frac{116}{3}

Therefore,

\frac{116}{3}

is the last term of the given AP.
Hence, the correct option is 4.

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