#### Question If the ratio of the sum of the first n terms of two APs is (7n + 1) : (4n + 27), then find the ratio of their 9^{th} terms.

#### Video Explanation

#### Explanatory Answer

Let the first term of the 1^{st} AP be a_{1}, and let its common difference be d_{1}.

Let the first term of the 2nd AP be b_{1}, and let its common difference be d_{2}.

Therefore, sum of the first n terms of 1^{st} AP = \\frac{n}{2})(2a_{1} + (n – 1)d_{1})

Sum of the first n terms of 2nd AP = \\frac{n}{2})(2b_{1} + (n – 1)d_{1})

The ratio of the sum up to first n terms of the two Aps is 7n + 1 : 4n + 27

i.e., \\frac {\frac {n} {2}\left ( {2{{a}_{1}+(n-1){d}_{1}}} \right )} {\frac {n} {2}\left ( {2{{b}_{1}+(n-1){d}_{2}}} \right )}) = \\frac{7n + 1}{4n + 27})

\\frac {\left ( {2{{a}_{1}+(n-1){d}_{1}}} \right )} {\left ( {2{{b}_{1}+(n-1){d}_{2}}} \right )}) = \\frac{7n + 1}{4n + 27}) ............ (1)

We have to compute a_{9} : b_{9}

a_{9} : b_{9} = 2a_{9} : 2b_{9}

2a_{9} = 2(a_{1} + 8d_{1}) = 2a_{1} + 16d_{1}

2b_{9} = 2(b_{1} + 8d_{2}) = 2b_{1} + 16d_{2}

Therefore, 2a_{9} : 2b_{9} = \\frac {2{{a}_{1}+16{d}_{1}}} {2{{b}_{1}+16{d}_{2}}}) ............ (2)

Comparing (1) and (2)

(n – 1) = 16 or n = 17

Therefore, a_{9} : b_{9} = \\frac{7(17) + 1}{4(17) + 27}) = \\frac{120}{95}) = \\frac{24}{19})