Question
If the $(a+1) t h, 7 t h$ and $(b+1)$ th terms of an $A P$ are in $G P$ with $a, 6, b$ being in $H P$, then 4 th term of this $A P$ is(a) $-\frac{7}{2}$(b) $\frac{7}{2}$(c) 0(d) 3
Step 1
Let's denote the common difference of the AP as $d$. Then, the $(a+1)$th term is $a_1 + ad$, the $7$th term is $a_1 + 6d$, and the $(b+1)$th term is $a_1 + bd$. Since these terms are in GP, we have \[(a_1 + 6d)^2 = (a_1 + ad)(a_1 + bd).\] Show more…
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