The ratio of sum of n terms of two AP is n+1 n+1

Text Solution

Solution : Let first term in first AP is `a_1` and common difference is `d_1`.<br> Let first term in second AP is `a_2` and common difference is `d_2`.<br> Then, we are given,<br> `S_1/S_2 = (2n+1)/(2n-1)`<br> `(n/2(2a_1+(n-1)d_1))/(n/2(2a_2+(n-1)d_2)) = (2n+1)/(2n-1)`<br> `=>(2a_1+(n-1)d_1)/(2a_2+(n-1)d_2) = (2n+1)/(2n-1)->(1)`<br> Now, we have to find ratio of `10th ` terms of both AP.<br> If we put `n = 19` in (1),<br> `(2a_1+18d_1)/(2a_2+18d_2) = 39/37`<br> `=> (a_1+9d_1)/(a_2+9d_2) = 39/37->(2)`<br> `10th` term of an AP is given by,<br> `T_10 = a+9d`<br> So, equation `(2)` gives us the required ratio that is `39:37`.<br>

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 $N=2n-1$.