Question

If $a,b,c$ are in $A.P$, then show that ${10}^{ax+10},{10}^{bx+10},{10}^{cx+10},x\ne 0$ are in $G.P$

Answer 1

a, b, c are in A.P

$\therefore b-a=c-b=d$ ($d=$common difference)….$\left(3\right)$

$\therefore \frac{{10}^{bx+10}}{{10}^{ax+10}}={10}^{bx+10-ax-10}$

$={10}^{x(b-a)}$ ……..$\left(1\right)$

$\therefore \frac{{10}^{cx+10}}{{10}^{bx+10}}={10}^{cx+10-bx-10}$

$={10}^{(c-b)x}$ ………..$\left(2\right)$

From $\left(1\right),\left(2\right)$ & $\left(3\right)$

${10}^{ax+10},{10}^{bx+10},{10}^{cx+10}$ are in G.P.