Question

the ratio of the sums pf first n yerms of two AP is (7n+1) : (4n+27). Find the ratio of mth terms of these AP.

Answer 1

Let the First AP be : AP₁

First term of AP₁ = a₁

Common Difference of AP₁ = d₁

Let the Second AP be : AP₂

First term of AP₂ = a₂

Common Difference of AP₂ = d₂

We know that Sum of 'n' terms of an AP is given by :

$S_n = \frac{n}{2}(2a + (n - 1)d)$

Given - Ratio of Sums of First 'n' terms of Two AP's as (7n + 1) : (4n + 27)

⇒ $\frac{S_n\;of\;AP_1}{S_n\;of\;AP_2} = \frac{7n + 1}{4n + 27}$

We need to find the ratio of mth terms :

We know that mth term is : a + (m - 1)d

$\Longrightarrow Ratio\;of\;m_t_h\;terms = \frac{a_1 + (m - 1)d_1}{a_2 + (m - 1)d_2}\\\\Multiplying\;and\;dividing\;with\;2\;we\;get\\\\\Longrightarrow \frac{2a_1 + 2(m - 1)d_1}{2a_2 + 2(m - 1)d_2}\\\\\Longrightarrow \frac{2a_1 + ([2m - 1] - 1)d_1}{2a_2 + ([2m - 1] - 1)d_2}\\\\Multiplying\;and\;dividing\;with\;\frac{(2m - 1)}{2}$

$\Longrightarrow \frac{\frac{(2m - 1)}{2}(2a_1 + ([2m - 1] - 1)d_1)}{\frac{(2m - 1)}{2}(2a_2 + ([2m - 1] - 1)d_2)}\\\\\Longrightarrow \frac{Sum\;of\;(2m - 1)\;terms\;of\;AP_1}{Sum\;of\;(2m - 1)\;terms\;of\;AP_2}\\ \\\Longrightarrow \frac{7(2m - 1) + 1}{4(2m - 1) + 27}\\\\\Longrightarrow \frac{14m - 6}{8m + 23}\\\\So,\;The\;Ratio\;of\;m_t_h\;Terms\; = \frac{14m - 6}{8m + 23}$