Question

if a,b,c are in arithmetic progression, then show that 10^ax+10,10^bx+10,10^cx+10, x not equals to zero, are in geometric progression.
Look carefully at q. no. 13

Answer 1

a,b,c are in ap
$a - b= b - c = d$
${10}^{ax + 10} \: {10}^{bx + 10} \: {10}^{cx + 10} \: are \: in \: gp \: iff \: common ratio \: is \: equal$
${10}^{ax + 10} \div {10}^{bx + 10} = {10}^{ax - bx} = {10}^{dx}$
${10}^{bx + 10} \div {10}^{cx + 10} = {10}^{bx - cx}= {10}^{dx}$
Common ratio are equal hence
${10}^{ax + 10} \: {10}^{bx + 10} \: {10}^{cx + 10} \: are \: in \: gp$