Question

If the sums of n terms of two arithmetic progressions are the ratio (2n + 3) : (6n + 5), then the ratio of their 13^th terms is ___________.

Answer 1

Let a₁ and d₂ represent the first term and common difference of first arithmetic progression.
Also, a₂ and d₂ represent the first term and common difference of second arithmetic progression then sum of n terms of first A.P
${\left(S_n\right)}_1=\frac{n}{2}\left\{2a_1+\left(n-1\right)d_1\right\}$
and sum of n terms of second A.P
${\left(S_n\right)}_2=\frac{n}{2}\left\{2a_2+\left(n-1\right)d_2\right\}$
Now, according to given condition
$$\begin{gathered} \frac{{\left(S_n\right)}_1}{{\left(S_n\right)}_2}=\frac{2n+3}{6n+5} \\ \mathrm{i}.\mathrm{e}\frac{2n+3}{6n+5}=\frac{{\displaystyle \frac{n}{2}}\left\{2a_1+\left(n-1\right)d_1\right\}}{{\displaystyle \frac{n}{2}}\left\{2a_2+\left(n-1\right)d_2\right\}} \\ \frac{2n+3}{6n+5}=\frac{a_1+{\displaystyle \frac{\left(n-1\right)}{2}{d}_1}}{a_2+{\displaystyle \frac{\left(n-1\right)}{2}}{d}_2} \end{gathered}$$
Since n^th term of any A.P is a + (n – 1)d
∴ 13^th term of first A.P is a₁ + 12d₁ and 13^th term of second A.P is a₂ + 12d₂
$$\begin{gathered} \mathrm{i}.\mathrm{e}\frac{n-1}{2}=12 \\ \mathrm{i}.\mathrm{e}\frac{n-1}{2}=12 \\ \mathrm{i}.\mathrm{e}n=24+1 \\ \mathrm{i}.\mathrm{e}n=25 \\ \therefore \frac{a_1+12d_1}{a_2+12d_2}=\left(\frac{2n+3}{6n+5}\right)atn=25 \\ =\frac{2\left(25\right)+3}{6\left(25\right)+5} \\ \therefore \frac{{\left(a_{13}\right)}_1}{{\left(a_{13}\right)}_2}=\frac{53}{155} \end{gathered}$$
i.e ratio of their 13^th terms is $\frac{53}{155}$.