Since a, b, c, d, are in G.P. Then $\frac{b}{a}$ = $\frac{c}{b}$ = $\frac{d}{c}$ ⇒ $\left(\frac{b}{a}\right)^n$ = $\left(\frac{c}{b}\right)^n$ = $\left(\frac{d}{c}\right)^n$ ⇒ $\frac{b^n}{a^n}$ = $\frac{c^n}{b^n}$ = $\frac{d^n}{c^n}$ ... (i) Adding 1 to (i), we obtain $\frac{b^n}{a^n}$ + 1 = $\frac{c^n}{b^n}$ + 1 = $\frac{d^n}{c^n}$ + 1 ⇒ $\frac{b^n + a^n}{a^n}$ = $\frac{c^n + b^n}{b^n}$ = $\frac{d^n + c^n}{c^n}$ Since $\frac{b^n + a^n}{a^n}$ = $\frac{c^n + b^n}{b^n}$ ⇒ $\frac{b^n + c^n}{a^n + b^n}$ = $\frac{b^n}{a^n}$ ... (ii) Now , $\frac{c^n + b^n}{b^n}$ = $\frac{d^n + c^n}{c^n}$ ⇒ $\frac{d^n + c^n}{c^n + b^n}$ = $\frac{c^n}{b^n}$ ... (iii) Then $\frac{b^n + c^n}{a^n + b^n}$ = $\frac{d^n + c^n}{b^n + c^n}$ [from (i), (ii) and (iii)] ⇒ $(b^n + c^n)^2$ = $(a^n + b^n)(a^n + c^n)$ Hence above equation shows that $(a^n + b^n), (b^n + c^n), (c^n + d^n)$ are in G. P.