Question

The ratio of the sum of n terms of 2 A.P.’s is (7n+1):(4n+27). Find the ratio of their 5 terms

Answer 1

$\bf \implies \green{{ \underline{given \div }}}$

$\rm \to \: ratio \: of \: sum \: of \: n \: terms \: of \: 2 \: a.p\\ \\ \rm \to \: (7n \: + \: 1) \ratio \: (4n \: + \: 27)$

$\bf \to \: \orange{\underline{to \: find \: the \: ratio \: of \: their \: 5th \: terms}}$

$\rm \to \: sum \: of \: n \: terms \: of \: an \: ap \\ \\ \rm \to \: s_{n} \: = \frac{n}{2} (2a \: + \: ( \: n \: - \: 1)d) \\ \\ \rm \to \: the \: ratio \: of \: sum \: of \: s_{n} \: and \: s_{n} {}^{1} \\ \\ \rm \to \: \frac{ s_{n}}{ s_{n} {}^{1} } = \frac{ \frac{n}{2}(2a \: + \: (n \: - \: 1)d) }{ \frac{n}{2}(2a {}^{1} + (n \: - \: 1)d {}^{1} ) } = \frac{7n \: + \: 1}{4n \: + \: 27} \\ \\ \rm \to \: \frac{ s_{n} }{ s_{n} {}^{1} } = \frac{2a \: + \: ( \: n - 1)d}{2a {}^{1} + (n - 1)d {}^{1} } = \frac{7n \: + \: 1}{4n \: + \: 27} \\ \\ \rm \to \: \frac{ s_{n} }{ s_{n} {}^{1} } = \frac{a \: + (\frac{n \: - \: 1}{2})d }{a {}^{1} + \frac{(n \: - \: 1)}{2} d {}^{1} } = \frac{7n \: + \: 1}{4n \: + \: 27} \\ \\ \rm \to \: we \: have \: to \: find \: 5th \: term \\ \\ \rm \to \frac{ t_{5} }{ t_{5} {}^{1} } = \frac{a \: + \: 4d}{a {}^{1} + 4d {}^{1} } \\ \\ \rm \to \: \frac{n \: - \: 1}{2} = 5 \\ \\ \rm \to \: n \: - 1 \: = 10 \\ \\ \rm \to \: n \: = \: 11 \\ \\ \rm \to \: \frac{ t_{5}}{ t_{5} {}^{1} } = \frac{7n \: + \: 1}{4n \: + \: 27} \\ \\ \rm \to \: \frac{ t_{5} }{ t_{5} {}^{1} } = \frac{7(11) + 1}{4(11) + 27} \\ \\ \rm \to \: \frac{ t_{5} }{ t_{5} {}^{1} } = \frac{78}{71}$