Geometric progression and logarithms

Question

I would like to ask you for some help, solving that:

'The sum of three members of a geometric progression ($a, aq, aq^2$) is $62$ and the sum of their decimal logarithms $lg$ is equal to $3$. $a$ and $q$ are positive numbers. Find ($a, aq, aq^2$).

I find it hard to figure out how to solve that and would be grateful if you can help me!

Answer 1

Partial answer:

Note that:

$\log(a)+\log(aq)+\log(aq^2)=\log(a\cdot aq\cdot aq^2)=\log(a^3q^3)=\log(aq)^3=3\log(aq)$

Therefore:

$\log(a)+\log(aq)+\log(aq^2)=3\implies3\log(aq)=3\implies\log(aq)=1\implies aq=10$

Therefore:

$a+aq+aq^2=62 \implies a+10+10q=62 \implies a=52-10q$

So now we have a system of two equations in two variables:

• $aq=10$
• $a=52-10q$

Which gives $a=52-\frac{100}{a} \implies a^2-52a+100=0 \implies (a-50)(a-2)=0$

Hence $[a=50]\vee[a=2]$

Hence $[q=\frac{10}{50}]\vee[q=\frac{10}{2}]$

Hence $[a,aq,aq^2=50,10,2]\vee[a,aq,aq^2=2,10,50]$