Question

if the ratio of the first n terms of two AP's is (7n+1) : (4n+27),find the ratio of their 9 th
terms

Answer 1

Sum of n terms=
$\frac{n}{2} (2a + (n - 1)d)$
nth term of an AP=
$a + (n - 1)d$
Now,
$\frac{ \frac{n}{2}(2a + (n - 1)d) }{ \frac{n}{2}(2b + (n - 1)e) } = \frac{7n + 1}{4n + 27}$
Here 'a' and 'b' are the first terms of the two APs and 'd' and 'e' are the common differences of the two APs.
$\frac{2a + (n - 1)d}{2b + (n - 1)e} = \frac{7n + 1}{4n + 27}$
Replace all n's with 2n-1
$\frac{2a + (2n - 2)d}{2b + (2n - 2)e } = \frac{7(2n - 1) + 1}{4(2n - 1) + 27}$
$\frac{a + (n - 1)d}{b + (n - 1)e} = \frac{14n - 6}{8n + 23}$
$\frac{nth \: term \: of \: 1st \: ap}{nth \: term \: of \: 2nd \: ap} = \frac{14n - 6}{4n + 23}$

Replace n with 9
$\frac{9th \: term \: of \: 1st \: ap}{ 9th \: term \: of \: 2nd \: ap} = \frac{24}{19}$

There you go.