Question

An A.P., a G.P. and H.P. have a and b for their first two terms, Show that their (n + 2) th terms will be in G.P. if $\frac{b^{2n+2}-a^{2n+2}}{ab(b^{2n}-a^{2n})}=\frac{n+1}{n}.$

Answer 1

d = b - a for A.P., r = b / a for G.P.
$T_{n+2}=a+(n+1)d=a+(n+1)(b-a)$
= -na + (n + 1)b for A.P. .....(1)
$T_{n+2}=a{r}^{n+1}=a.\frac{b^{n+1}}{a^{n+1}}=\frac{b^{n+1}}{a^n}$ for G.P. ...(2)
For getting $T_{n+2}$ foe H.P., replace a by $\frac{1}{a},$ and b by $\frac{1}{b}$ in $T_{n+2}$
$\therefore T_{n+2}$ for H.P. = reciprocal of $-n\left(\frac{1}{a}\right)+(n+1)\frac{1}{b}$
or of $\frac{(n+1)a-nb}{ab}$
$\therefore T_{n+2}=\frac{ab}{(n+1)a-nb}$ for H.P. ....(3)
The above three terms are themselves in G.P.
$\therefore \left(\frac{b^{n+1}}{a^n}\right)=[-na+(n+1\left)b\right].\frac{ab}{(n+1)a-nb}$
$\therefore dfrac{b}^{2n+1}{a}^{2n+1}=\frac{-na+(n+1)b}{(n+1)a-nb}$ Cancelled ab
Cross multiply
$(n+1)[a{b}^{2n+1}-b{a}^{2n+1}]=n(b^{2n+2}-a^{2n+2})$
or $\frac{n+1}{n}=\frac{b^{2n+2}-a^{2n+2}}{ab(b^{2n}-a^{2n})}$