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The sum of $n$ terms of two arithmetic series are in the ratio of $(7n+ 1) : (4n+ 27)$. We have to find the ratio of their $n$-th term. � I tried to find the ratio by using the formula of summation of A.P.

But it becomes too long due to many variables that is $a_1,a_2,d_1,d_2$

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    Hmm,but I wish you googled it once. – 2017-01-26

4 Answers 4

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It is actually quite simple. Let $a_1$ and $a_1'$ denote the first terms of the first and second progressions with their common differences $d$ and $d'$ respectively. We thus get $$\frac{S_1}{S_2} = \frac {0.5n (2a_1 +(n-1)d)}{0.5n (2a_1' +(n-1)d')} = \frac {2a_1+(n-1)d}{2a_1' +(n-1)d'} = \frac {7n+1}{4n+27} $$

The ratio of the $n$th term of the two AP's can be thus calculated as $$\frac{a_n}{a_n'} = \frac {a_1 +(n-1)d}{a_1'+(n-1)d'} = \frac {2a_1 +((2n-1)-1)d}{2a_1' + ((2n-1)-1)d'} $$ $$=\frac {S_{2n-1}}{S_{2n-1}'} = \frac {14n-6}{8n+23} $$ Hope it helps.

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    Nice answer. +1 – 2017-01-26
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Hint: The sum of $nth$ term of a AP is given by $$\frac{n}{2}[a+a_n]$$.The ratio of first term can easily be found out by putting $n=1$.So it is easy to find out the ratio of $nth$ term using this.

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We have $$\frac{7n+1}{4n+27}=\dfrac{\dfrac n2\{2a_1+(n-1)d_1\}}{\dfrac n2\{2a_2+(n-1)d_2\}}=\dfrac{a_1+\dfrac{(n-1)}2\cdot d_1}{a_2+\dfrac{(n-1)}2\cdot d_2}$$

Replace $\dfrac{n-1}2$ with $m-1\iff n=2m-1$ to find the ratio of their $m$th term.

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The sum to $n$ terms of an AP is of the form $\lambda(An^2+Bn)$, (where $A,B, \lambda$ are constants) although when taking ratios for the sums for two APs, $n, \lambda$ cancel out, giving $\frac{^1An+^1B}{^2An+^2B}$. The ratio of the $n$-th terms can be derived by considering the ratio of the differences between the sum to $n$ terms and sum to $(n-1)$ terms respectively, based on the given ratio of sums but first adjusting for the point described above.

$$\begin{align} \frac {^1u_n}{^2u_n}&=\frac{^1S_n-^1S_{n-1}}{^2S_n-^2S_{n-1}}\\ &=\frac{\color{blue}n[7n+1]-\color{blue}{(n-1)}[7(n-1)+1]}{\color{blue}n[4n+27]-\color{blue}{(n-1)}[4(n-1)+27]}\\ &=\frac{7[n^2-(n-1)^2]+1}{4[n^2-(n-1)^2]+27}\\ &=\frac{7(2n-1)+1}{4(2n-1)+27} \color{lightgrey}{=\frac{7N+1}{4N+27}=\frac {^1S_{N}}{^2S_N}}\\ &=\color{red}{\frac {14n-6}{8n+23}}\end{align}$$

Note that, as also pointed out in other solutions posted earlier, this is the same as the ratio of the sum to $N$ terms of the two APs where $N=2n-1$.