Let the first three terms 2, p and q, with q 2, of a G.P.

Question

Let the first three terms 2, p and q, with $q
e 2$, of a G.P. be respectively the ${7}^{},{8}^{}$ and ${13}^{}$ terms of an A.P. If the ${5}^{}$ term of the G.P. is the ${n}^{}$ term of the A.P., then $n$ is equal to:

Accepted Answer

$$\begin{aligned}
& \text { Let } p=2 r, q=2 r^2 \
& T_7=2, T_8=2 r, T_{13}=2 r^2 \
& d=2 r-2=2(r-1) \
& 2 r^2=T_7+6 d=2+6(2)(r-1)=12 r-10 \
& \Rightarrow r^2-6 r+5=0 \
& \Rightarrow(r-1)(r-5)=0 \
& \therefore r=1,5 \
& r=1 \text { (rejected) as } q \neq 2 \
& \therefore r=5
\end{aligned}$$

$$5^{\text {th }}$$ term of G.P $=2 . r^4=2.5^4$

Let $$1^{\text {st }}$$ term of A.P $b a=a, d=8$

$$2=a+(6)(8) \Rightarrow a=-46$$

$$\mathrm{n}^{\text {th }}$$ term of A.P $=-46+(n-1) 8=8 n-54$

$$\begin{aligned}
& 2.5^4=8 n-54 \
& \Rightarrow 1250+54=8 n \
& \Rightarrow n=\frac{1304}{8}=163
\end{aligned}$$

Solutions · 1