If for an $A . P$ $T_{3} = 18$ and $T_{7} = 30$ , then $S_{17}$ is equal to

612

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622

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306

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510

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Solution

The correct option is B $510$
$a_{3} = a + 2 \times d = 18$ $a + 6 = 18$
$a_{7} = a + 6 d = 30$ $a = 12$
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$4 d = 12$
$d = 3$
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$S_{17} = \frac{17}{2} [12 + 16 \times 3] = \frac{17}{2} \times 60 = 510$ .

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