Question

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Q. If the ratio of the sum of the first n terms of 2 A.P are (7n+1) : (4n+27) , then find the ratio of both A.P's 9th term

Answer 1

Let the First term of the First A.P be : a₁

Let the Common difference of the First A.P be : d₁

Let the First term of the Second A.P be : a₂

Let the Common difference of the Second A.P be : d₂

We know that :

$\bigstar\;\;\mathsf{Sum\;of\;n\;terms\;in\;an\;A.P\;is\;given\;by : \boxed{\mathsf{\dfrac{n}{2}[2a + (n - 1)d]}}}$

$\mathsf{Given - Ratio\;of\;sum\;of\;the\;first\;n\;terms\;of\;two\;A.P's\;is : \dfrac{7n + 1}{4n + 27}}$

$\implies \mathsf{\dfrac{\dfrac{n}{2}[2a_1 + (n - 1)d_1]}{\dfrac{n}{2}[2a_2 + (n - 1)d_2]} = \dfrac{7n + 1}{4n + 27}}$

$\implies \mathsf{\dfrac{[2a_1 + (n - 1)d_1]}{[2a_2 + (n - 1)d_2]} = \dfrac{7n + 1}{4n + 27}}$

$\implies \mathsf{\dfrac{\bigg[a_1 + \bigg(\dfrac{n - 1}{2}\bigg)d_1\bigg]}{\bigg[a_2 + \bigg(\dfrac{n - 1}{2}\bigg)d_2\bigg]} = \dfrac{7n + 1}{4n + 27}\;-----\;[1]}$

The Question is to find the ratio of both A.P's 9th term's

$\bigstar\;\;\mathsf{We\;know\;that,\;n^{th}\;term\;in\;an\;A.P\;is\;given\;by : \boxed{\mathsf{T_n = a + (n - 1)d}}}$

$\implies \mathsf{Ratio\;of\;9^{th}\;term's\;of\;Two\;A.P's\;will\;be : \dfrac{a_1 + (9 - 1)d_1}{a_2 + (9 - 1)d_2}}$

$\implies \mathsf{Ratio\;of\;9^{th}\;term's\;of\;Two\;A.P's\;will\;be : \dfrac{a_1 + (8)d_1}{a_2 + (8)d_2}}$

Now, Comparing LHS of Equation [1] & the ratio of Two A.P's 9th terms :

$\mathsf{\dfrac{\bigg[a_1 + \bigg(\dfrac{n - 1}{2}\bigg)d_1\bigg]}{\bigg[a_2 + \bigg(\dfrac{n - 1}{2}\bigg)d_2\bigg]} = \dfrac{a_1 + (8)d_1}{a_2 + (8)d_2}}$

$\mathsf{We\;can\;notice\;that : \bigg(\dfrac{n - 1}{2}\bigg) = 8}$

$\mathsf{\implies (n - 1) = 16}$

$\mathsf{\implies n = 17}$

Substituting n = 17 in Equation [1], We get :

$\implies \mathsf{\dfrac{\bigg[a_1 + \bigg(\dfrac{17 - 1}{2}\bigg)d_1\bigg]}{\bigg[a_2 + \bigg(\dfrac{17 - 1}{2}\bigg)d_2\bigg]} = \dfrac{7(17) + 1}{4(17) + 27}}$

$\implies \mathsf{\dfrac{\bigg[a_1 + \bigg(\dfrac{16}{2}\bigg)d_1\bigg]}{\bigg[a_2 + \bigg(\dfrac{16}{2}\bigg)d_2\bigg]} = \dfrac{119 + 1}{68 + 27}}$

$\implies \mathsf{\dfrac{a_1 +8d_1}{a_2 + 8d_2} = \dfrac{120}{95}}$

$\implies \mathsf{\dfrac{a_1 +8d_1}{a_2 + 8d_2} = \dfrac{24}{19}}$

$\implies \mathsf{\dfrac{First\;A.P - Ninth\;term}{Second\;A.P - Ninth\;term} = \dfrac{24}{19}}$

Answer - The Ratio of both A.P's Ninth term's is 24 : 19