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Mathematics

Fourth and seventh terms of a G.P. are 118 and −1486\dfrac{1}{18} \text{ and } -\dfrac{1}{486} respectively. Find the G.P.

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Answer

Let first term of the G.P. be a and it's common ratio be r.

Given,

⇒ a4 = 118\dfrac{1}{18}

⇒ ar3 = 118\dfrac{1}{18} ……..(i)

Also,

⇒ a7 = −1486-\dfrac{1}{486}

⇒ ar6 = −1486-\dfrac{1}{486} ……..(ii)

Dividing (ii) by (i) we get,

⇒ar6ar3=−1486118⇒r3=−18486⇒r3=−127⇒r3=(−13)3⇒r=−13.\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{-\dfrac{1}{486}}{\dfrac{1}{18}} \\[1em] \Rightarrow r^3 = -\dfrac{18}{486} \\[1em] \Rightarrow r^3 = -\dfrac{1}{27} \\[1em] \Rightarrow r^3 = \Big(-\dfrac{1}{3}\Big)^3 \\[1em] \Rightarrow r = -\dfrac{1}{3}.

Substituting value of r in (i) we get,

⇒a×(−13)3=118⇒a×−127=118⇒a=−2718=−32.\Rightarrow a \times \Big(-\dfrac{1}{3}\Big)^3 = \dfrac{1}{18} \\[1em] \Rightarrow a \times -\dfrac{1}{27} = \dfrac{1}{18} \\[1em] \Rightarrow a = -\dfrac{27}{18} = -\dfrac{3}{2}.

⇒ a2 = ar

= −32×(−13)-\dfrac{3}{2} \times \Big(-\dfrac{1}{3}\Big)

= 12\dfrac{1}{2}.

⇒ a3 = ar2

= −32×(−13)2-\dfrac{3}{2} \times \Big(-\dfrac{1}{3}\Big)^2

= −32×19-\dfrac{3}{2} \times \dfrac{1}{9}

= −16.-\dfrac{1}{6}.

G.P. = −32,12,−16,118...........-\dfrac{3}{2}, \dfrac{1}{2}, -\dfrac{1}{6}, \dfrac{1}{18} ………..

Hence, G.P. = −32,12,−16,118...........-\dfrac{3}{2}, \dfrac{1}{2}, -\dfrac{1}{6}, \dfrac{1}{18} ………..

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