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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?

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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.

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