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Let the first three terms $2, p$ and $q$, with $q \neq 2$, of a G.P. be respectively the $7^{\text {th }}, 8^{\text {th }}$ and $13^{\text {th }}$ terms of an A.P. If the $5^{\text {th }}$ term of the G.P. is the $n^{\text {th }}$ term of the A.P., then $n$ is equal to:
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The correct answer is:
163
$p^2=2 q$
$2=a+6 d$ ...(i)
$\mathrm{p}=\mathrm{a}+7 \mathrm{~d}$ ...(ii)
$q=a+12 d$ ...(iii)
$\mathrm{p}-2=\mathrm{d}$ ((ii) - (i))
$q-p=5 d$ ((iii) - (ii))
$q-p=5 d$
$\begin{aligned} & q-p=5(p-2) \\ & q=6 p-10 \\ & p^2=2(6 p-10) \\ & p^2-12 p+20=0 \\ & p=10,2 \\ & p=10 ; q=50 \\ & d=8 \\ & a=-46 \\ & 2,10,50,250,1250 \\ & a^4=a+(n-1) d \\ & 1250=-46+(n-1) 8 \\ & n=163\end{aligned}$
$2=a+6 d$ ...(i)
$\mathrm{p}=\mathrm{a}+7 \mathrm{~d}$ ...(ii)
$q=a+12 d$ ...(iii)
$\mathrm{p}-2=\mathrm{d}$ ((ii) - (i))
$q-p=5 d$ ((iii) - (ii))
$q-p=5 d$
$\begin{aligned} & q-p=5(p-2) \\ & q=6 p-10 \\ & p^2=2(6 p-10) \\ & p^2-12 p+20=0 \\ & p=10,2 \\ & p=10 ; q=50 \\ & d=8 \\ & a=-46 \\ & 2,10,50,250,1250 \\ & a^4=a+(n-1) d \\ & 1250=-46+(n-1) 8 \\ & n=163\end{aligned}$
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