MCQ
The common difference of an $A.P.$ whose first term is unity and whose second, tenth and thirty fourth terms are in $G.P.$, is
- A$\frac{1}{5}$
- ✓$\frac{1}{3}$
- C$\frac{1}{6}$
- D$\frac{1}{9}$
$\therefore {T_2} = a + d,\,\,$
${T_{10}} = a + 9d,$
${T_{34}} = a + 33d$
$\therefore {(a + 9d)^2} = (a + d)(a + 33d)$
==> ${a^2} + 81{d^2} + 18ad = {a^2} + ad + 33ad + 33{d^2}$
Put $a = 1$
$ \Rightarrow 1 + 81{d^2} + 18d = 1 + d + 33d + 33{d^2}$
==> $48{d^2} - 16d = 0$
$ \Rightarrow 16d(3d - 1) = 0$
==> $d = 0,\,\,d = 1/3$.
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