Question

Ratio of sum of n terms of two AP's = (7n + 1) : (4n + 27). Find the ratio of mth term.

Answer 1

Given ratio of sum of n terms of two AP's = (7n + 1) : (4n + 27)
$\frac{n}{2}(2a+(n-1\left)d\right):\frac{n}{2}(2a^{{}^{\prime }}+(n-1\left)d^{{}^{\prime }}\right)=(7n+1):(4n+27)(2a+(n-1\left)d\right):(2a^{{}^{\prime }}+(n-1\left)d^{{}^{\prime }}\right)=(7n+1):(4n+27)\to \left(1\right)$
Let's consider the ratio these two AP's mth terms as $a_m:a_m^{{}^{\prime }}\to \left(2\right)$
Recall the nth term of AP formula, $a_n=a+(n-1)d$
Hence equation (2) becomes,
$a_m:a_m^{{}^{\prime }}=a+(m-1)d:a^{{}^{\prime }}+(m-1)d^{{}^{\prime }}$
On multiplying by 2, we get
$a_m=a_m^{{}^{\prime }}=[2a+2(m-1\left)d\right]:[2a^{{}^{\prime }}+2(m-1\left)d^{{}^{\prime }}\right]=[2a+\{(2m-1)-1\left\}d\right]:[2a^{{}^{\prime }}+\{(2m-1)-1\left\}{d}^{{}^{\prime }}\right]=S_{2m-1}:S_{2m-1}^{{}^{\prime }}=\left[7\right(2m-1)+1]:\left[4\right(2m-1)+27]\left[from\right(1\left)\right]=[14m-7+1]:[8m-4+27]=[14m-6]:[8m+23]$
Thus the ratio of mth terms of two AP's is [14m - 6] : [8m + 23]