Question

Let a1, a2, a3, a4, a5 be a G.P. of positive real numbers such that the A.M. of a2 and a4 is 117 and the G.M. of a2 and a4 is 108. Then the A.M. of a1 and a5 is : (1) 108 (2) 117 (3) 144.5 (4) 145.5

Answer 1

$\underline{\underline{\mathfrak{\Large{Solution : }}}}$



$\textsf{Let,} \\ \\ \sf \implies a_1 \: = \: ar^{-2} \\ \\ \sf \implies a_2 \: = \: ar^{-1} \\ \\ \sf \implies a_3 \: = \: a\\ \\ \sf \implies a_4 \: = \: ar \\ \\ \sf \implies a_5 \: = \: ar^{2}$





$\underline{\textsf{According to question , }} \\ \\ \sf \implies A.M. \: of \: a_2 \: and \: a_4 \: = \: 117 \\ \\ \sf \implies \dfrac{a_2 \: + \: a_4 }{2} \: = \: 117 \\ \\ \sf \implies \dfrac{ar^{-1} \: + \: ar}{2} \: = \: 117 \\ \\ \sf \implies ar^{-1} \: + \: ar \: = \: 117 \: \times \: 2 \\ \\ \\ \sf \implies a(r^{-1} \: + \: r ) \: = \: 234 \qquad...(1)$






$\textsf{Again,} \\ \\ \sf \implies G.M. \: of \: a_2 \: and \: a_4 \: = \: 108 \\ \\ \sf \implies \sqrt{a_2 \: \times \: a_4 } \: = \: 108 \\ \\ \sf \implies \sqrt{ar^{-1} \: \times \: ar } \: = \: 108 \\ \\ \sf \implies a^{(1 \: + \: 1)} \: \times \: r^{(-1 \: + \: 1)} \: = \: (108)^2 \\ \\ \sf \implies a^2 \: \times \: r^0 \: = \: (108)^2 \\ \\ \sf \implies a^2 \: = \: ( 108)^2 \\ \\ \sf \implies a \: = \: 108 \qquad...(2)$





$\textsf{Plug the value of (2) in (1), } \\ \\ \sf \implies a(r^{-1} \: + \: r ) \: = \: 234 \\ \\ \sf \implies 108(r^{-1} \: + \: r ) \: = \: 234 \\ \\ \sf \implies \left( \dfrac{1}{r} \: + \: r \right) \: = \: \dfrac{234}{108} \\ \\ \sf \implies \dfrac{ {r}^{2} \: + \: 1}{r} \: = \: \dfrac{13}{6} \\ \\ \sf \implies 6 {r}^{2} \: + \: 6 \: = \: 13r \\ \\ \sf \implies6 {r}^{2} \: - \: 13r \: + \: 6 \: = \: 0 \\ \\ \sf \implies 6 {r}^{2} \: - \: 9r \: - \: 4r \: + \: 6 \: = \: 0 \\ \\ \sf \implies3r(2r \: - \: 3 ) \: - \: 2(2r \: - \: 3) \: = \: 0 \\ \\ \sf \implies(2r \: - \: 3)(3r \: - \: 2) \: = \: 0$





$\underline{\textsf{By Zero Product Rule : }} \\ \\ \sf \implies 2r \: - \: 3 \: = \: 0 \quad \implies 3r \: - \: 2 \: = \: 0 \\ \\ \sf \implies 2r \: = \: 3 \: \: \: \: \: \: \: \: \: \: \quad \implies 3r \: = \: 2 \\ \\ \sf \implies \: \: \therefore \: r \: = \: \dfrac{3}{2} \: or \: \dfrac{2}{3}$





$\textsf{Now,} \\ \\ \sf Let, A.M. \: of \: a_1 \: and \: a_5 \: is \: x . \\ \\ \sf \implies \dfrac{a_1 \: + \: a_5 }{2} \: = \: x \\ \\ \sf \implies \dfrac{ar^{-2} \: + \: ar^2 }{2} \: = \: x \\ \\ \sf \implies \dfrac{a( r^{-2} \: + \: r^2)}{2} \: = \: x \\ \\ \sf \implies \dfrac{108( r^{-2} \: + \: r^2)}{2} \: = \: x \\ \\ \sf \implies 54(r^{-2} \: + \: r^{2} ) \: = \: x$





$\sf When , \: r \: = \: \dfrac{3}{2} : \\ \\ \sf \implies 54 \left\{ \left( \dfrac{3}{2} \right)^{( - 2)} \: + \: \left(\dfrac{3}{2} \right)^2 \right \}\: = \: x \\ \\ \sf \implies 54 \left \{ \left( \dfrac{2}{3} \right)^{2} \: + \: \left( \dfrac{3}{2} \right)^{2} \right \} \: = \: x \\ \\ \sf \implies 54 \left( \dfrac{4}{9} \: + \: \dfrac{9}{4} \right) \: = \: x \\ \\ \sf \implies 54 \left( \dfrac{16 \: + \: 81}{36} \right) \: = \: x \\ \\ \sf \implies 54 \: \times \: \dfrac{97}{36} \: = \: x \\ \\ \sf \: \therefore \: \: x \: = \: 145.5$






$\sf When , \: r \: = \: \dfrac{2}{3} : \\ \\ \sf \implies 54 \left\{ \left( \dfrac{2}{3} \right)^{( - 2)} \: + \: \left(\dfrac{2}{3} \right)^2 \right \}\: = \: x \\ \\ \sf \implies 54 \left \{ \left( \dfrac{3}{2} \right)^{2} \: + \: \left( \dfrac{2}{3} \right)^{2} \right \} \: = \: x \\ \\ \sf \implies 54 \left( \dfrac{9}{4} \: + \: \dfrac{4}{9} \right) \: = \: x \\ \\ \sf \implies 54 \left( \dfrac{81\: + \: 16}{36} \right) \: = \: x \\ \\ \sf \implies 54 \: \times \: \dfrac{97}{36} \: = \: x \\ \\ \sf \: \therefore \: \: x \: = \: 145.5$




$\sf Hence , the \: A.M. \: of \: a_1 \: and \: a_5 \: is \: 145.5$

Answer 2

$\underline{\underline{\mathfrak{\Large{Solution : }}}}$



$\textsf{Let,} \\ \\ \sf \implies a_1 \: = \: ar^{-2} \\ \\ \sf \implies a_2 \: = \: ar^{-1} \\ \\ \sf \implies a_3 \: = \: a\\ \\ \sf \implies a_4 \: = \: ar \\ \\ \sf \implies a_5 \: = \: ar^{2}$





$\underline{\textsf{According to question , }} \\ \\ \sf \implies A.M. \: of \: a_2 \: and \: a_4 \: = \: 117 \\ \\ \sf \implies \dfrac{a_2 \: + \: a_4 }{2} \: = \: 117 \\ \\ \sf \implies \dfrac{ar^{-1} \: + \: ar}{2} \: = \: 117 \\ \\ \sf \implies ar^{-1} \: + \: ar \: = \: 117 \: \times \: 2 \\ \\ \\ \sf \implies a(r^{-1} \: + \: r ) \: = \: 234 \qquad...(1)$






$\textsf{Again,} \\ \\ \sf \implies G.M. \: of \: a_2 \: and \: a_4 \: = \: 108 \\ \\ \sf \implies \sqrt{a_2 \: \times \: a_4 } \: = \: 108 \\ \\ \sf \implies \sqrt{ar^{-1} \: \times \: ar } \: = \: 108 \\ \\ \sf \implies a^{(1 \: + \: 1)} \: \times \: r^{(-1 \: + \: 1)} \: = \: (108)^2 \\ \\ \sf \implies a^2 \: \times \: r^0 \: = \: (108)^2 \\ \\ \sf \implies a^2 \: = \: ( 108)^2 \\ \\ \sf \implies a \: = \: 108 \qquad...(2)$





$\textsf{Plug the value of (2) in (1), } \\ \\ \sf \implies a(r^{-1} \: + \: r ) \: = \: 234 \\ \\ \sf \implies 108(r^{-1} \: + \: r ) \: = \: 234 \\ \\ \sf \implies \left( \dfrac{1}{r} \: + \: r \right) \: = \: \dfrac{234}{108} \\ \\ \sf \implies \dfrac{ {r}^{2} \: + \: 1}{r} \: = \: \dfrac{13}{6} \\ \\ \sf \implies 6 {r}^{2} \: + \: 6 \: = \: 13r \\ \\ \sf \implies6 {r}^{2} \: - \: 13r \: + \: 6 \: = \: 0 \\ \\ \sf \implies 6 {r}^{2} \: - \: 9r \: - \: 4r \: + \: 6 \: = \: 0 \\ \\ \sf \implies3r(2r \: - \: 3 ) \: - \: 2(2r \: - \: 3) \: = \: 0 \\ \\ \sf \implies(2r \: - \: 3)(3r \: - \: 2) \: = \: 0$





$\underline{\textsf{By Zero Product Rule : }} \\ \\ \sf \implies 2r \: - \: 3 \: = \: 0 \quad \implies 3r \: - \: 2 \: = \: 0 \\ \\ \sf \implies 2r \: = \: 3 \: \: \: \: \: \: \: \: \: \: \quad \implies 3r \: = \: 2 \\ \\ \sf \implies \: \: \therefore \: r \: = \: \dfrac{3}{2} \: or \: \dfrac{2}{3}$





$\textsf{Now,} \\ \\ \sf Let, A.M. \: of \: a_1 \: and \: a_5 \: is \: x . \\ \\ \sf \implies \dfrac{a_1 \: + \: a_5 }{2} \: = \: x \\ \\ \sf \implies \dfrac{ar^{-2} \: + \: ar^2 }{2} \: = \: x \\ \\ \sf \implies \dfrac{a( r^{-2} \: + \: r^2)}{2} \: = \: x \\ \\ \sf \implies \dfrac{108( r^{-2} \: + \: r^2)}{2} \: = \: x \\ \\ \sf \implies 54(r^{-2} \: + \: r^{2} ) \: = \: x$





$\sf When , \: r \: = \: \dfrac{3}{2} : \\ \\ \sf \implies 54 \left\{ \left( \dfrac{3}{2} \right)^{( - 2)} \: + \: \left(\dfrac{3}{2} \right)^2 \right \}\: = \: x \\ \\ \sf \implies 54 \left \{ \left( \dfrac{2}{3} \right)^{2} \: + \: \left( \dfrac{3}{2} \right)^{2} \right \} \: = \: x \\ \\ \sf \implies 54 \left( \dfrac{4}{9} \: + \: \dfrac{9}{4} \right) \: = \: x \\ \\ \sf \implies 54 \left( \dfrac{16 \: + \: 81}{36} \right) \: = \: x \\ \\ \sf \implies 54 \: \times \: \dfrac{97}{36} \: = \: x \\ \\ \sf \: \therefore \: \: x \: = \: 145.5$






$\sf When , \: r \: = \: \dfrac{2}{3} : \\ \\ \sf \implies 54 \left\{ \left( \dfrac{2}{3} \right)^{( - 2)} \: + \: \left(\dfrac{2}{3} \right)^2 \right \}\: = \: x \\ \\ \sf \implies 54 \left \{ \left( \dfrac{3}{2} \right)^{2} \: + \: \left( \dfrac{2}{3} \right)^{2} \right \} \: = \: x \\ \\ \sf \implies 54 \left( \dfrac{9}{4} \: + \: \dfrac{4}{9} \right) \: = \: x \\ \\ \sf \implies 54 \left( \dfrac{81\: + \: 16}{36} \right) \: = \: x \\ \\ \sf \implies 54 \: \times \: \dfrac{97}{36} \: = \: x \\ \\ \sf \: \therefore \: \: x \: = \: 145.5$




$\sf Hence , the \: A.M. \: of \: a_1 \: and \: a_5 \: is \: 145.5$