# Sum of squares in geometric progression

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?

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 $b_{n-1}b_{n+1}=b_{n}^{2}$.

• After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$ – user167045 Mar 2 '15 at 5:04
• @Garvil I don't see any leftover terms. – Kugelblitz Mar 2 '15 at 5:10
• Yeah, sorry. My bad. – user167045 Mar 2 '15 at 5:12
• No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible. – Kugelblitz Mar 2 '15 at 5:13
• Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :) – Janko Bracic Mar 2 '15 at 5:27

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: $$(10)^2 -(5)^2 = (10+5)(10-5) = 75.$$

• Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil – Kugelblitz Mar 2 '15 at 5:09
• 20 terms cancel out; remaining five are the required five. – Kugelblitz Mar 2 '15 at 5:11