Question

the ratio of 7th to the 3rd term of an ap is 12 is 25 find the ratio of 13th to the 4th term ​

Answer 1

Correction in the Question: The ratio of the 7th term to the 3rd term of an AP is 12 is to 5 and the ratio of the 13th term to the 4th term is?

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.

ATQ,

$\Longrightarrow \sf \dfrac{a_7}{a_3} = \dfrac{12}{5}$

$\Longrightarrow \sf \dfrac{a + 6d}{a + 2d} = \dfrac{12}{5}$

Cross multiplying we get,

$\Longrightarrow \sf 5(a + 6d) = 12 (a + 2d)$

$\Longrightarrow \sf 5a + 30d = 12a + 24d$

$\Longrightarrow \sf 5a - 12a = 24d - 30d$

$\Longrightarrow \sf -7a = -6d$

$\Longrightarrow \sf 7a = 6d$

$\Longrightarrow \sf a = \dfrac{6d}{7}$

We have to find,

$\Longrightarrow \sf \dfrac{a_{13}}{a_4}$

$\Longrightarrow \sf \dfrac{a + 12d}{a + 3d}$

Substituting the value of 'a' we get,

$\Longrightarrow \sf \dfrac{\left(\dfrac{6d}{7} + 12d\right)}{\left(\dfrac{6d}{7} + 3d}\right)$

$\Longrightarrow \sf \dfrac{\left(\dfrac{6d + 84d}{7}\right)}{\left(\dfrac{6d + 21d}{7}\right)}$

7, The Denominator gets cancelled.

$\Longrightarrow \sf \dfrac{6d + 84d}{6d + 21d}$

$\Longrightarrow \sf \dfrac{90d}{27d}$

Dividing them we get,

$\Longrightarrow \sf \dfrac{10d}{3d}$

Cancelling 'd' we get.

$\Longrightarrow \sf \dfrac{10}{3}$

Hence, the ratio of the 13th to the 4th term of the given AP is 10:3.