does there exist a GP containing 27, 8, 12 and 3 of its terms if it exists how many such progression are possible??
Answers
Answered by
5
Please find the answer to your question
If possible let for a G. P.
Tp = 27 = ARp-1 …………. (1)
Tq = 8 = ARq-1 ………… (2)
Tr = 12 = ARr-1 ………… (3)
From (1) and (2)
R p-q = 27/8 ⇒ Rp – q = (3/2)3 ……….. (4)
From (2) and (3) ;
Rq – r = 8/12 ⇒ Rq – r = (3/2) -1 ………….. (5)
From (4) and (5)
R = 3/2 ; p-q = 3; q – r = -1
p – 2q + r = 4; p, q, r ∈ N ………… (6)
As there can be infinite natural numbers for p, q, and r to satisfy equation (6)
∴ There can be infinite G. P’ s.
If possible let for a G. P.
Tp = 27 = ARp-1 …………. (1)
Tq = 8 = ARq-1 ………… (2)
Tr = 12 = ARr-1 ………… (3)
From (1) and (2)
R p-q = 27/8 ⇒ Rp – q = (3/2)3 ……….. (4)
From (2) and (3) ;
Rq – r = 8/12 ⇒ Rq – r = (3/2) -1 ………….. (5)
From (4) and (5)
R = 3/2 ; p-q = 3; q – r = -1
p – 2q + r = 4; p, q, r ∈ N ………… (6)
As there can be infinite natural numbers for p, q, and r to satisfy equation (6)
∴ There can be infinite G. P’ s.
Similar questions