Question

The ratio of sum to n terms of two AP's is 8n+1/7n+3 for every n ∈ N. Find the ratio of their 7th term and nth term.

Answer 1

Dear Student,

Solution:

Let the first AP has first term = a1 , common difference = d1

Sum of n terms Sn = n/2(2a1+(n-1)d1)

Let the second AP has first term = a2 , common difference = d2

Sum of n terms Sn = n/2(2a2+(n-1)d2)

According to the question: Ratio of sum of both AP

$\frac{\frac{n}{2} (2a_{1} +(n-1)d_{1} )}{\frac{n}{2} (2a_{2} +(n-1)d_{2} )} =\frac{8n+1}{7n+3}$

$\frac{2a_{1}+(n-1)d_{1}}{2a_{2}+(n-1)d_{2}} =\frac{8n+1}{7n+3}$

To find the ratio of mth term, we have to replace n by (2m-1) in the above expression.

Replacing n by (2*7-1) ,i.e. 13 on both sides

$\frac{2a_{1}+(13-1)d_{1}}{2a_{2}+(13-1)d_{2}} =\frac{8*13+1}{7*13+3}$

$\frac{2a_{1}+12d_{1}}{2a_{2}+12d_{2}} =\frac{105}{94} \\ \\ \frac{a_{1}+6d_{1}}{a_{2}+6d_{2}} =\frac{105}{94}$

So, The ration of 7th term of both APs = 105:94

hope it helps you.