Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

3 Answers

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 $15$.

share|improve this answer
add comment

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?

share|improve this answer
    
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
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.