Question

An A.P., a G.P., and an H.P. have $a$ and $b$ for their first two terms. Their $(n+2{)}^{th}$ terms will be in G.P. if $\frac{b^{2n+2}-a^{2n+2}}{ab(b^{2n}-a^{2n})}$

• A. $\frac{n}{n+1}$
• B. $\frac{n+1}{n}$
• C. $\frac{n+1}{n+2}$
• D. $\frac{2n+1}{n+2}$

Answer 1

The correct option is B

$\frac{n+1}{n}$

The $(n+2{)}^{th}$ term of A.P.
$x_1=a+(n+1)(b-a)$
The $(n+2{)}^{th}$ term of G.P.
$x_2=a{\left(\frac{b}{a}\right)}^{n+1}$
the $(n+2{)}^{th}$ term of H.P.
$x_3=\frac{1}{\frac{1}{a}+(n+1)(\frac{1}{b}-\frac{1}{a})}$
Now ,$x_1,x_2,x_{3}areinG.P.if{x}_2^2=x_1{x}_3$ that is
$if{a}^2{\left(\frac{b}{a}\right)}^{2n+2}=\frac{a+(n+1)(b-a)}{\frac{1}{a}+(n+1)(\frac{1}{b}-\frac{1}{a})}or\frac{b^{2n+2}}{a^{2n}}=\frac{(n+1)b-na}{b+(n+1)(a-b)}ab\therefore \frac{b^{2n+2}-a^{2n+2}}{ab(b^{2n}-a^{2n})}=\frac{n+1}{n}$