Sum of squares in geometric progression

Question

In the geometric progression $b_1, b_2, b_3,\ldots, b_1+b_3+b_5=10$ and $b_2+b_4=5$. Find the sum of the squares of the first five terms.

If you solve for the first term and the common ratio, you will get nasty radicals and then squaring and adding is not a good option. How do I do it then?

Answer 1

Hint: $(b_1+b_3+b_5)^2-(b_2+b_4)^2$.

More suggestions: $(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz$, $(x+y)^2=x^2+2xy+y^2$ (this you know, I guess), now use the fact that in geometric progression one has $b_{n-1}b_{n+1}=b_{n}^{2}$.

Answer 2

Let $b_1=a$, and the common ratio be $r$. Given that $$a+ar^2+ar^4=10 \qquad\text{and}\qquad ar+ar^3=5.$$ What is the value of $$(a)^2 +(ar)^2 +(ar^2)^2 +(ar^3)^2 +(ar^4)^2?$$ Realize that: $$(a+ar^2+ar^4)^2 -(ar+ar^3)^2$$ simplifies to the required sum. On substituting: $$(10)^2 -(5)^2 = (10+5)(10-5) = 75.$$