Find the common ratio, if the sum of the first $8$ terms in a geometric progression is equal to $17$ times the sum of its first $4$ terms

Question

Find the common ratio, if the sum of the first $8$ terms in a geometric progression (GP) is equal to $17$ times the sum of its first $4$ terms.

So far I have got $$a+ar+ar^2+ar^3+ar^4+ar^5+ar^6+ar^7=17(a+ar+ar^2+ar^3).$$ How do I proceed without using the sum of GP formula? Thank you.

Answer 1

Hint:

$$a+ar+ar^2+ar^3+ar^4+ar^5+ar^6+ar^7=17(a+ar+ar^2+ar^3)$$

$$a(1+r+r^2+r^3)+ar^4(1+r+r^2+r^3)=17a(1+r+r^2+r^3)$$

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Hint 2:

After the above step, we can gather everything onto the left-hand side and then factor:

$$ar^4(1+r+r^2+r^3)-16a(1+r+r^2+r^3)=0$$ $$a(r^4-16)(1+r+r^2+r^3)=0$$

Can you see what to do from here?

Answer 2

$a+ar+ar^2+ar^3+ar^4+ar^5+ar^6+ar^7=17(a+ar+ar^2+ar^3)$

$a(1+r+r^2+r^3)+a^4(1+r+r^2+r^3)=17a(1+r+r^2+r^3)$

Cancel out $a(1+r+r^2+r^3)$ to get:

$1+a^3=17$

or, $a^3=16$

Thus, $a=2\sqrt[3]2$