Question

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

Answer 1

$a,b,c$ are in $AP$ : -

$b-a=c-b$

$\Rightarrow bx-ax=cx-bx$

$\Rightarrow (bx+10)-(ax+10)=(cx+10)-(bx+10)$

${10}^{(bx+10)-(ax+10)}={10}^{(cx+10)-(bx+10)}$

$\Rightarrow {10}^{(bx+10)}=\sqrt{\left({10}^{an+10}\right)\left({10}^{bx+10}\right)}$

$\therefore$${10}^{an+10},{10}^{bx+10},{10}^{cx+10}$ are in $GP$.