Question

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

Answer 1

Answer:

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Step-by-step explanation:

Product of three terms = 216

(a/r) ⋅ a ⋅ a r = 216

a3 = 63

a = 6

Sum of their product in pairs = 156

(a/r) ⋅ a + a ⋅ ar + ar ⋅ (a/r) = 156

a2 / r + a2 r + a2 = 156

a2 [ (1/r) + r + 1 ] = 156

a² [ (1+r²+r)/r] = 156

a² (r²+r+1)/r = 156

(6²/r)(r²+r+1) = 156

(r²+r+1)/r = 156/36

(r²+r+1)/r = 13/3

3(r²+r+1) = 13 r

3r² + 3r + 3 - 13r = 0

3r² - 10r + 3 = 0

(3r - 1)(r - 3) = 0

3r - 1 = 0

3r = 1

r = 1/3

r - 3 = 0

r = 3

If a = 6, then r = 1/3

a/r = 6/(1/3) ==> 18

a = 6

ar = 6(1/3) ==> 2

If a = 6, then r = 1/3

a/r = 6/3 ==> 2

a = 6

ar = 6(3) ==> 18

Hence the required three terms are 18, 6 and 2 or 2, 6, 18.