Question

sum of n terms of 2 APs is (5n+3):(3n+4)then ratio of thier 17th tem is

Answer 1

$\bf \to\underline{For \: the \: first \: ap} \\ \sf \: \: \: \: \: \: \: \: \: \bullet \:First \: term = a \\ \sf \: \: \: \: \: \: \: \: \: \bullet \:Common \: difference \: = d \\ \sf \: \: \: \: \: \: \: \: \: \bullet \: Sum \: of \: terms = s_{n} = \frac{n}{2} 2a + (n - 1)d \\ \: \sf \: \: \: \: \: \: \: \: \: \bullet \: {n}^{th} term = a_{n} = a + (n - 1)d\\ \\ \to\bf \underline{For \: the \: second \: term} \\ \sf \: \: \: \: \: \: \: \: \: \bullet \:First \: term = A \\ \sf \: \: \: \: \: \: \: \: \: \bullet \:Common \: difference = D \\ \sf \: \: \: \: \: \: \: \: \: \bullet \: s_{n} = \frac{n}{2} 2A + (n - 1)D \\ \sf \: \: \: \: \: \: \: \: \: \bullet \: A_{n} = A + (n - 1)D$

$\\ \\ \blue{ \implies} \bf We \: need \: to \: find \: ratio \: of \: 17th \: term.$

$\implies \sf \: \: \: \: \: \: \: \: \dfrac{17th \: term \: of \: 1st \: ap}{17th \: term \: of \: 2nd \: ap} \\ \\ \: \: \: \: \: \: \: \: = \sf \dfrac{a_{17} \: of \: 1st \: ap}{a_{17} \: of \: 2nd \: ap} \\ \\ \: \: \: \: \: \: \: \: \sf = \dfrac{a + (17 - 1)d}{A + (1 7 - 1)D} \\ \\ \: \: \: \: \: \: \: \: \sf = \frac{a + 16d}{A + 16D}$

$\\ \bf \because \underline{Given \: that}$

$\implies\sf \dfrac{Sum \: of \: nth \: terms \: of \: 1st \: ap}{Sum \: of \: nth \: term \: of \: 2nd \: term} = \dfrac{5n + 3}{3n + 4} \\ \\ \sf \: \: \: \: \: \: \: \: \implies \dfrac{ \dfrac{n}{2} 2a + (n - 1)d}{ \dfrac{n}{2} 2A + (n - 1)D} \\ \\ \: \: \: \: \: \: \: \: \implies \sf \: \dfrac{ a + (n - 1)d}{ A + (n - 1)D} ......eq(1)$

$\\ \therefore \bf \underline{We \: have \: to \: find} \: \to \frac{a + 16d}{A + 16D}$

$\implies \dfrac{n - 1}{2} = 16 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: n - 1 = 32 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{n = 33}$

➸ Putting n = 33 in eq(1),

$\: \: \: \: \: \: \sf \dfrac{ a + \dfrac{(33 - 1)d}{2} }{A + \dfrac{(33 - 1)}{2} D} = \dfrac{5(33) + 3}{3(33) + 4} \\ \\ \: \: \: \: \: \sf \dfrac{a + \dfrac{32}{2} d}{A + \dfrac{32}{2}D } = \dfrac{168}{103} \\ \\ \large \implies \boxed {\boxed {\tt \blue {\sf \dfrac{a + 16 d}{A + 16D } = \dfrac{168}{103} }}} \\ \\ \\ \sf \bf \implies \therefore \: \dfrac{17th \: term \: of \: 1st \: term}{17th \: term \: of \: 2nd \: term} = \dfrac{168}{103}$

Hence, ratio of their 17th term is 168:103.