Sum of first $5$ terms of an A.P. is one fourth of the sum of next five terms. If the first term is $2,$ then the common difference of the A.P. is

6

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- 6

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3

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None of these

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Solution

The correct option is A $- 6$
let the A.P. be $a, a + d, a + 2 d . . . . . . .$
We have $T_{1} + T_{2} + T_{3} + T_{4} + T_{5} = \frac{1}{4} [T_{6} + T_{7} + T_{8} + T_{9} + T_{10}]$
$∵$ Sum $= \frac{n}{2} ($ first term $+$ last term $)$
Then $\frac{5}{2} (T_{1} + T_{5}) = \frac{1}{4} \times \frac{5}{2} (T_{6} + T_{10})$
$\Rightarrow \frac{5}{2} (a + (a + 4 d)) = \frac{5}{8} [(a + 5 d) (a + 9 d)]$
$\Rightarrow 2 a + 4 d = \frac{1}{4} [2 a + 14 d]$
$\Rightarrow 4 a + 8 d = a + 7 d$
$\Rightarrow d = - 3 a$ .... $(a = 2)$
$\Rightarrow d = - 3 \times 2 = - 6$

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