Question

For two different APs, the ratio of the sums of the first n terms is $\frac{7n+1}{4n+27}$. What is the ratio of the $m^{th}$ terms of the two APs? [4 MARKS]

Answer 1

Formula: 1 Mark each
Application: 1 Mark
Answer: 1 Mark

Let the first terms and common differences of the two APs be $a_1,a_2$ and $d_1,d_2$. We have:

$\frac{S_{1,n}}{S_{2,n}}=\frac{7n+1}{4n+27}$

$\Rightarrow \frac{\frac{n}{2}(2a_1+(n-1\left)d_1\right)}{\frac{n}{2}(2a_2+(n-1\left)d_2\right)}=\frac{7n+1}{4n+27}$

$(\because S_n=\frac{n}{2}\times (2a+(n-1)d\left)\right)$

$\Rightarrow \frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{7n+1}{4n+27}$

Note that the above equality is true for every natural number n. Now, we replace $n\to 2m-1$.

$\Rightarrow \frac{2a_1+(2m-2)d_1}{2a_2+(2m-2)d_2}=\frac{7(2m-1)+1}{4(2m-1)+27}$

$\Rightarrow \frac{a_1+(m-1)d_1}{a_2+(m-1)d_2}=\frac{14m-6}{8m+23}$

$\Rightarrow \frac{T_{1,m}}{T_{2,m}}=\frac{14m-6}{8m+23}$

$(\because T_n=a+(n-1\left)d\right)$

The replacement $n\to 2m-1$ converted the expression on the left to the ratio of the $m^{th}$ terms of the two series.