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:
Solution
<p>$$\begin{aligned}
& \text { Let } p=2 r, q=2 r^2 \\
& T_7=2, T_8=2 r, T_{13}=2 r^2 \\
& d=2 r-2=2(r-1) \\
& 2 r^2=T_7+6 d=2+6(2)(r-1)=12 r-10 \\
& \Rightarrow r^2-6 r+5=0 \\
& \Rightarrow(r-1)(r-5)=0 \\
& \therefore r=1,5 \\
& r=1 \text { (rejected) as } q \neq 2 \\
& \therefore r=5
\end{aligned}$$</p>
<p>$5^{\text {th }}$ term of G.P $=2 . r^4=2.5^4$</p>
<p>Let $1^{\text {st }}$ term of A.P $b a=a, d=8$</p>
<p>$2=a+(6)(8) \Rightarrow a=-46$</p>
<p>$\mathrm{n}^{\text {th }}$ term of A.P $=-46+(n-1) 8=8 n-54$</p>
<p>$$\begin{aligned}
& 2.5^4=8 n-54 \\
& \Rightarrow 1250+54=8 n \\
& \Rightarrow n=\frac{1304}{8}=163
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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