Question

The ratio of 11 th term and 18 th term is 2:3 then ratio of sum of 1st 5 term to sum of 1st 10 terms is

Answer 1

Question: The ratio of the 11th term and the 18th term of an AP is 2:3 then the ratio of the sum of the first 5 terms to the sum of the first 10 terms is?

ATQ,

$\Longrightarrow \sf \dfrac{a_{11}}{a_{18}} = \dfrac{2}{3}$

We know that a term of an AP can be expressed as the first term plus the product of the common difference and the position of the previous term.

$\Longrightarrow \sf \dfrac{a + 10d}{a + 17d} = \dfrac{2}{3}$

$\sf \Longrightarrow 3(a+10d) = 2(a+17d)$

$\sf \Longrightarrow 3a+30d = 2a+34d$

$\sf \Longrightarrow 3a-2a = 34d-30d$

$\sf \Longrightarrow a = 4d$

ATQ, We have to find,

$\Longrightarrow \sf \dfrac{S_5}{S_{10}}$

$\Longrightarrow \sf \dfrac{\frac{n}{2}\left(2a+(n-1)d\right)}{\frac{n}{2}\left(2a+(n-1)d\right)}$

Substituting 'a' with '4d' we get,

$\Longrightarrow \sf \dfrac{\frac{5}{2}\left(2(4d)+(5-1)d\right)}{\frac{10}{2}\left(2(4d)+(10-1)d\right)}$

$\Longrightarrow \sf \dfrac{\frac{5}{2} \left(8d+4d\right)}{5\left(8d+9d\right)}$

$\Longrightarrow \sf \dfrac{\frac{5}{2} \left(12d\right)}{5\left(17d\right)}$

Taking 2 as common out from the numerator,

$\Longrightarrow \sf \dfrac{\frac{5}{2} \times 2 \left(6d\right)}{5\left(17d\right)}$

$\Longrightarrow \sf \dfrac{5\left(6d\right)}{5\left(17d\right)}$

Cancelling 5.

$\Longrightarrow \sf \dfrac{6d}{17d}$

Cancelling 'd'.

$\Longrightarrow \sf \dfrac{6}{17}$

$\Longrightarrow \ \boxed{\sf \dfrac{S_5}{S_{10}} = \dfrac{6}{17}}$

Hence, the sum of the first 5 terms to the sum of the first 10 terms is 6:17.