If $a_{3} = 9$ and $a_{7} = 21$ . Find $S_{5}$

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Solution

The correct option is B 45
Using the formula,
$a_{n} = a + (n - 1) d$
So, $a_{3} = a + (3 - 1) d$
$a = a + 2 d$ ......... (1)
$a_{7} = a + (7 - 1) d$
$21 = a + 6 d$ ........ (2)
Subtracting equation (1) & (2)
$a = a + 2 d$ ........ (1)
$21 = a = b d$ ....... (2)
(-) (-) (-)
------------------
$- 20 = - 4 d$
$d = 5$
Put the value of d in equation (1)
$a = a + 2 (5)$
$a = - 1$
The sum of first n term is given by $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$S_{5} = \frac{5}{2} [2 (- 1) + (5 - 1) 5]$
$= \frac{5}{2} (- 2 + 20)$
$= 2.5 (18)$
$S_{5} = 45$

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S_{5} = 15

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