# Find the total of the sum of the first five terms of the arithmetic series and the sum of the first three terms of the geometric series.

Following information is available regarding the series:

1. The second term of the geometric series is the same as the fourth term of the arithmetic series.
2. The $7^{th}$ term of the arithmetic series is the same as the $3^{rd}$ term of the geometric series.
3. The first term of the geometric series exceeds the first term of the arithmetic series by 64/3.
4. The sum of the first three terms of the arithmetic series, $A_3$ and the sum of the first two terms of the geometric series, $G_2$ are related by the formula $A_3 + G_2 + 21 = 0$.

To find: The total of the sum of the first 5 terms of the arithmetic series and the sum of the first 3 terms of the geometric series.

My approach so far:

Let's assume that the first term of AP be $a$ and the common difference be $d$, the first term of GP be $b$ and the common ratio be $r$. Then we have following: $$br = a + 3d$$ $$br^2 = a + 6d$$ $$b-a=\frac{64}{3}$$ $$a+d+b+br+7=0$$

We have to find: $5a+10d+b+br+br^2$.

So far I have got: $$a+d+b+br+7+br^2=a+6d \implies b+br+br^2=5d-7$$ Hence, $$5a+10d+b+br+br^2=5a+10d+(5d-7)=5(a+3d)-7=5br-7$$

I am stuck at this point as I am not able to think of any way this can be solved without involving higher powers of the variables.

How can we proceed from here?

Hint: From $A_3+G_2+21=0$ we obtain $3a+3d+b+br+21=0$.
There are four equations in four variables and one of them contains $r^2$. So, you have to solve a quadratic equation as there is no nice cancellation.