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# If the product of three consecutive terms in G.P. is 216 and sum of their products in pairs is 156, find them.

Answer :

Let the first term of G.P be .

⇒ second term = a and third term = ar.

(where, r is the common ratio)

∵ sum of three terms is 156

⇒ a^{3} = 216

⇒ a = 6. ……..(1)

Also,sum of their product in pairs is 216.

………(2)

Substituting (1) in (2), we get-

⇒ 36(1 + r + r^{2}) = 156r

⇒ 36 + 36r + 36r^{2} = 156r

⇒ 36 -120r + 36r^{2} = 0

⇒ 12(3r^{2}-10r + 3) = 0

⇒ 3r^{2}-10r + 3 = 0

⇒ 3r^{2}-9r-1r + 3 = 0

⇒ 3r(r-3) -1(r-3) = 0

⇒ (3r-1)(r-3) = 0

Now, G.P will be-

⇒18, 6, 2 or 2, 6, 18 are the three consecutive terms.

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