Question

If the ratio of the sum of the first n terms of two A.Ps (7n+ 1) : ( 4n + 27) then find the ratio of their 9th terms.
PLEASE GIVE THE ANSWER STEP BY STEP.
AND PLEASE DON'T POST THE SILLY ANSWER.
PLEASE HELP ME.

Answer 1

Given that ratio of the sum of the first n terms of two AP's = (7n + 1) : (4n + 27).

$= \ \textgreater \ \frac{2a + (n - 1) * d}{2a' + (n - 1) * d'} = \frac{7n + 1}{4n + 27}$    ----------- (1)

Ratio of their 9th terms:

$= \ \textgreater \ \frac{a + 8d}{a' + 8d}$

$= \ \textgreater \ \frac{2a + 16d}{2a' + 16d' }$

$= \ \textgreater \ \frac{2a + (17 - 1) * d}{2a' + (17 - 1) * d'}$

Therefore n = 17.

Substitute n = 17 in (1), we get

$= \ \textgreater \ \frac{7(17) + 1}{4(17) + 27}$

$= \ \textgreater \ \frac{120}{95}$

$= \ \textgreater \ \frac{24}{19}$



Hope this helps!