# Arithmetic and Geometric Series Problem

In an arithmetical progression, the ratio of the 2nd term to the 4th is 11:13 and the sum of the first five terms is 30. Find the sum of thirty terms. (Ans: 367.5) Some help, please?

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Maybe in this case we could let the third term be $t$. Why third term? Because the information given seems symmetrical about the third term. Let $d$ as usual be the common difference.

The five terms are $p$, $p\pm d$, and $p\pm 2d$. They add up to $30$, so $p=6$.

The fourth term and the second are in the ratio $13:11$. Perhaps it is obvious then that they are $6.5$ and $5.5$, giving common difference $d=0.5$. Or else we can write $$\frac{6+d}{6-d}=\frac{11}{13}$$ and use algebra to solve for $d$.

The first term is $6-(2)(0.5)$, which is $5$. Now it is maybe safest to use a remembered formula for the sum of the first $n$ terms.

Or else calculate the $30$-th term, which is $5+(29)(0.5)$.
The sum of the first $30$ terms is then the average of the ends, times $30$, or equivalently the sum of the ends, times $15$.

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Let's summarize what you know. I will call the series $x_1, x_2 = x_1 + d,x_3 = x_1 + 2d, x_4 = x_1 + 3d, x_5 = x_1 + 4d, ...$.

We know that $\dfrac{x_2}{x_4} = \dfrac{11}{13}$. Alternatively, this means that $\dfrac{x_1 + d}{x_1 + 3d} = \dfrac{11}{13}$.

And we know that $x_1 + x_2 + ... + x_5 = 5 x_1 + (d + 2d + 3d + 4d) = 5 x_1 + 10d = 30$.

These two equations have two unknowns and are linear. Does that help?

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Oh yes, why didn't I see the connection? Thanks a lot! – Sophia Jun 28 '11 at 13:15
In your first line, it seems all the $x_0$s should be $x_1$s – Ross Millikan Jun 28 '11 at 13:15
@Oh: thanks Ross - I changed my initial form to match the numbering of the terms, but not everything. I will change it immediately. – mixedmath Jun 28 '11 at 13:19

In an AP, the terms are $a,a+d,a+2d,a+3d,\dots$ and the sum of the first $n$ terms is $(2a+(n-1)d)(n/2)$.

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