the sum of three numbers in a gp is 56.If we subtract 1,7,21
Answers
Answer:
Step-by-step explanation:
Let the three numbers in G.P. be a, ar, and ar2.
From the given condition, a + ar + ar2 = 56
⇒ a (1 + r + r2) = 56
rightwards double arrow a space equals fraction numerator 56 over denominator 1 plus r plus r squared end fraction space space space space... space left parenthesis 1 right parenthesis
a – 1, ar – 7, ar2 – 21 forms an A.P.
∴(ar – 7) – (a – 1) = (ar2 – 21) – (ar – 7)
⇒ ar – a – 6 = ar2 – ar – 14
⇒ar2 – 2ar + a = 8
⇒ar2 – ar – ar + a = 8
⇒a(r2 + 1 – 2r) = 8
⇒ a (r – 1)2 = 8 … (2)
rightwards double arrow fraction numerator 56 over denominator 1 plus r plus r squared end fraction open parentheses r minus 1 close parentheses squared equals 8 space space space space space space space space space space open square brackets U sin g space left parenthesis 1 right parenthesis close square brackets
⇒7(r2 – 2r + 1) = 1 + r + r2
⇒7r2 – 14 r + 7 – 1 – r – r2 = 0
⇒ 6r2 – 15r + 6 = 0
⇒ 6r2 – 12r – 3r + 6 = 0
⇒ 6r (r – 2) – 3 (r – 2) = 0
⇒ (6r – 3) (r – 2) = 0
therefore space r space equals space 2 comma 1 half
When r = 2, a = 8
When r space equals space 1 half space comma space a space equals space 32
Therefore, when r = 2, the three numbers in G.P. are 8, 16, and 32.
When r space equals space 1 half, the three numbers in G.P. are 32, 16, and 8.
Thus, in either case, the three required numbers are 8, 16, and 32.