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(B) Let the first term $a = 1$ and the common difference be $d$.The terms of the arithmetic progression are $t_n = a + (n-1)d$.Thus,$t_2 = 1 + d$,$t_{

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$1/3$

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(B) Let the first term $a = 1$ and the common difference be $d$.The terms of the arithmetic progression are $t_n = a + (n-1)d$.Thus,$t_2 = 1 + d$,$t_{10} = 1 + 9d$,and $t_{34} = 1 + 33d$.Since $t_2, t_{10}, t_{34}$ are in geometric progression,we have $(t_{10})^2 = t_2 	imes t_{34}$.$(1 + 9d)^2 = (1 + d)(1 + 33d)$.$1 + 18d + 81d^2 = 1 + 33d + d + 33d^2$.$1 + 18d + 81d^2 = 1 + 34d + 33d^2$.$81d^2 - 33d^2 + 18d - 34d = 0$.$48d^2 - 16d = 0$.$16d(3d - 1) = 0$.Since $d 
eq 0$ for a non-trivial progression,$3d - 1 = 0$,which gives $d = 1/3$.

The first term of an arithmetic progression is $1$. If the second,tenth,and thirty-fourth terms form a geometric progression,then the common difference of the arithmetic progression is:

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