Question

in G.P the sum of three number is 13. and the the sum of square of those three number is 91.then find the those numbers of GP

Answer 1

Let the 3 terms of G.P are a, ar, ar2
Then according to question,
Sum of the given terms = 13⇒ a + ar + ar2 = 13
⇒ a (1 + r + r2) = 13
⇒ a = 131 + r + r2 ...............(1)


Also it is given that sum of the squares of the given numbers = 91⇒ a2 + a2r2 + a2r4 = 91
⇒ a2 (1 + r2 + r4) = 91
⇒ a2 (1 + r2 + r4 + r2 − r2) = 91
⇒ a2 {(1 + 2r2 + r4) − r2} = 91
⇒ a2 {(1 + r2)2 − r2} = 91
⇒ a2 {(1 + r2 + r) (1 + r2 − r)} = 91
⇒ a2 = 91(1 + r + r2) (1 − r + r2) ................(2)


Putting the value of a from (1) in (2)

⇒ (131 + r + r2 )2 = 91(1 + r + r2) (1 − r + r2)
⇒ 169(1 + r + r2 ) (1 + r + r2 ) = 91(1 + r + r2) (1 − r + r2)
⇒ 13(1 + r + r2 ) = 7(1 − r + r2)
⇒ 13 − 13r + 13r2 = 7 + 7r + 7r2
⇒ 6r2 − 20r + 6 = 0
⇒ 3r2 − 10r + 3 = 0
⇒ 3r2 − 9r − r + 3 = 0
⇒ 3r (r − 3) − 1 (r − 3) = 0
⇒(r − 3) (3r − 1) = 0


Either, r − 3 = 0 or 3r − 1 = 0Either, r = 3 or r= 13



CASE (i). When r = 3 Then a = 131 + 3 + 9 [putting r = 3 in (1)] a = 1313 = 1Then the required numbers are: a, ar, ar2 = 1, 1×3, 1×9 = 1, 3, 9



CASE(ii). When r = 13 Then a = 131 + 13 + 19 a = 13139 = 13 × 913 = 9Then the required numbers are: a, ar, ar2 = 9, 9 ×13, 9 ×19 = 9, 3, 1 ( ans)