# In a geometric sequence, the sum of the first six terms is 9 times the sum of the first three terms. If the first term is f5, what is the third term?

I solved this question and my answer was 20, but apparently, it's supposed to be -25/2 and I don't understand how that works out.

$$S_6=9(S_3)$$, $$t_1=5$$

$$\frac{1-r^6}{1-r} = 9t_1(\frac{1-r^3}{1-r})$$

$$(1-r^3)(1-r^3)= 9(1-r^3)$$

$$(1+r^3)=9$$

$$r^3=8,\ 2^3$$

$$r=2$$

therefore: $$t_3= 5(2)^{3-1}$$

$$t_3=20$$

Your answer is correct. Here it is worked out again in a different manner:

Remembering a geometric sequence is in the form $$a,ax,ax^2,ax^3,\cdots$$ we are told the sum of the first six terms is equal to 9 times the sum of the first three terms.

$$a+ax+ax^2+ax^3+ax^4+ax^5 = 9(a+ax+ax^2)$$

Letting $$a=5$$ and moving to one side, this gives

$$5x^5+5x^4+5x^3-40x^2-40x-40=0$$

Factoring a bit:

$$5x^3(x^2+x+1) - 40(x^2+x+1)=0$$

$$(5x^3-40)(x^2+x+1)=0$$

Now... one of the solutions can be seen from the first factor as being when $$5x^3=40$$ and so $$x=\sqrt[3]{8}=2$$

The remaining four solutions can all be found with the quadratic formula and can be seen to all be complex.

So, we have found that for the sequence to be comprised only of real entries, the ratio between terms must be $$2$$.

The first term being $$5$$, that leaves the second term being $$10$$ and the third term being $$20$$.

• Thank you so much! This really helped me out! Commented Feb 15, 2019 at 1:20

Your work is correct, as we can manually check that $$r = 2$$ works: $$S_6 = 5 + 10 + 20 + 40 + 80 + 160 = 315 \\ 9S_3 = 9(5 + 10 + 20) = 315$$

• Thank you so much. I was really doubting myself because of the answer key, but I guess I can prove it to her if needed. You really helped me out :) Commented Feb 15, 2019 at 1:04