The common difference of an A.P. whose first | vedclass

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(B) Let the first term of the $A.P.$ be $a = 1$ and the common difference be $d$.The terms are given by $T_n = a + (n-1)d$.Thus,$T_2 = 1 + d$,$T_{10}

Vedclass · Accepted Answer

$\frac{1}{3}$

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(B) Let the first term of the $A.P.$ be $a = 1$ and the common difference be $d$.The terms are given by $T_n = a + (n-1)d$.Thus,$T_2 = 1 + d$,$T_{10} = 1 + 9d$,and $T_{34} = 1 + 33d$.Since $T_2, T_{10}, T_{34}$ are in $G.P.$,we have $(T_{10})^2 = T_2 	imes T_{34}$.$(1 + 9d)^2 = (1 + d)(1 + 33d)$.$1 + 81d^2 + 18d = 1 + 33d + d + 33d^2$.$1 + 81d^2 + 18d = 1 + 34d + 33d^2$.$48d^2 - 16d = 0$.$16d(3d - 1) = 0$.This gives $d = 0$ or $d = \frac{1}{3}$.Since the common difference is typically non-zero for such problems,the value is $\frac{1}{3}$.

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

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