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Find the $S_{15}$ for an AP whose nth term is given by $a_{n}$ $= 3 + 4 n$ .

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420

525

630

675

answer is B.

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Detailed Solution

Given,

a_{n} = 3 + 4 n

. . . (1)
We know that nth term of an AP is given by,

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

, where

a

is the first term and

d

the common difference of an AP.
On substituting

n = 1, 2, 3, . .

successively in (1), we get,

a_{1} = 3 + 4 \times 1 = 7

a_{2} = 3 + 4 \times 2 = 11

a_{3} = 3 + 4 \times 3 = 15

and so on.
Therefore, the sequence is

7, 11, 15, . . . . .

It is clear that the sequence obtained is an AP with first term,

a = 7

and common difference,

d = 4

.
Also, the sum of first

n

terms is given by,

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

⇒

S_{15} = \frac{15}{2} [2 a + (15 - 1) d]

\Rightarrow S_{15} = \frac{15}{2} [2 \times 7 + 14 \times 4]

= 15 \times 35

= 525

Hence, the correct option is 2.

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