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

420

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525

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630

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675

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Solution

The correct option is A $525$
We know that nth term, $a_{n} = a + (n - 1) d$
& Sum of first $n$ terms, $S_{n} = \frac{n}{2} (2 a + (n - 1) d)$ , where $a$ & $d$ are the first term amd common difference of an AP.
Since, $a_{n} = 3 + 4 n$ ...(1)
On putting $n = 1, 2, 3, . .$ in successively, 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..
Thus the sequence is $7, 11, 15, . . . . .$
Clearly it is an AP with first term $a = 7$ and common difference $d = 11 - 7 = 4$
Now, Required sum $S_{15} = \frac{15}{2} [2 a + (15 - 1) d]$
$\Rightarrow S_{15} = \frac{15}{2} [2 \times 7 + 14 \times 4]$
$= 15 \times 35$
$= 525$

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