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STD 11 - 8. sequence and series — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 8. sequence and series1 Mark
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
    $\frac{1}{9}$
✓

Answer

Correct option: B.
$\frac{1}{3}$
b
(b) First term of an $A.P. = 1$, let Common difference $= d$

$\therefore {T_2} = a + d,\,\,$

${T_{10}} = a + 9d,$

${T_{34}} = a + 33d$

$\therefore {(a + 9d)^2} = (a + d)(a + 33d)$

==> ${a^2} + 81{d^2} + 18ad = {a^2} + ad + 33ad + 33{d^2}$

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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