Question

The sums of n terms of two arithmetic series are in the ratio $2n+3:6n+5$, then the ratio of their ${13}^{th}$ terms is?

• A. $53:155$
• B. $27:77$
• C. $29:83$
• D. $31:89$

Answer 1

The correct option is A $53:155$

We are given that $\frac{(S_n{)}_1}{(S_n{)}_2}=\frac{2n+3}{6n+5}$ .......$\left(1\right)$

$\to$ Sum of n terms of an AP

$S_n=n/2[2a+(n-1\left)d\right]$

$S{O}_2=\frac{(S_n{)}_1}{(S_n{)}_2}=\frac{n/2[2a_1+(n-1\left)d_1\right]}{n/2[2a_2+(n-1\left)d_2\right]}$ .........$\left(2\right)$

$\to a,a_1$ & $a_2$ be the first term

$d,d_1$ & $d_2$ be the common difference

& n be the number of terms in an A.P.

From $\left(1\right)$ & $\left(2\right)$,

$\frac{2n+3}{6n+5}=\frac{n/2[2a_1+(n-1\left)d_1\right]}{n/2[2a_2+(n-1\left)d_2\right]}=\frac{2a_1-(n-1)d_1}{2a_2+(n-1)d_2}$

$\frac{2n+3}{6n+5}=\frac{a_1+\left(\right(n-1\left)/2\right)d_1}{a_2+\left(\right(n-1\left)/2\right)d_2}$ .........$\left(3\right)$

$\to$ Now $n^{th}$ term in an A.P. $=a+(n-1)d$

So, $\frac{(a_{13}{)}_1}{(a_{13}{)}_2}=\frac{a_1+12d_1}{a_2+12d_2}$ .........$\left(4\right)$

$\to$ In the equation $\left(3\right)$ we can put $n=25$

So, we get equation $\left(4\right)$

So, $\frac{2n+3}{6n+5}=\frac{a_1+12d_1}{a_2+12d_2}=\frac{(a_{13}{)}_1}{(a_{13}{)}_2}$, $n=25$

$\frac{(a_{13}{)}_1}{(a_{13}{)}_2}=\frac{2\left(25\right)+3}{6\left(25\right)+5}=\frac{53}{155}$

$\frac{(a_{13}{)}_1}{(a_{13}{)}_2}=\frac{53}{155}$.