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The common difference d of the A.P. in which ${T_{7}}^{_{}} = 9$
and $T_{1} T_{2} T_{7}$ is least is

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7/3

5/4

33/20

33/2

answer is C.

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Detailed Solution

$\begin{array}{l} T_{1} T_{2} T_{7} = (T_{7} - 6 d) (T_{7} - 5 d) T_{7} \\ = 9 (9 - 6 d) (9 - 5 d) \end{array}$

$= 729 (d - \frac{3}{2}) (d - \frac{9}{5})$

But $(x - a) (x - b)$ is least when $x = \frac{1}{2} (a + b)$

$∴ T_{1} T_{2} T_{7}$ is least when $d = \frac{1}{2} (\frac{3}{2} + \frac{9}{5}) = \frac{33}{20}$

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