Find three numbers in Gp. such that their sum is 21 and sum of their squares is 189.
Answers
Step-by-step explanation:
Q.) Find three consecutive no. s such that theur sum is 21 and sum of their squares is 149.
Ans) let first no. Be X
Second no. Be X+1
Third no. Be X+2...
Then by calculation, X= 6
X+1= 7
X+2= 8
Then u can cheak it, A. T. Q,
(X)²+(X+1)²+(X+2)²= 149.
6²+7²+8²=149.
36+49+64=149.
Plz!! Mark brainliest!!!
Answer:
Step-by-step explanation:
Suppose the 3 no.s in GP are : (a/r), a, (ar)............... (A)
∵ their sum is 21
∴ a( 1/r + 1 + r ) = 21 ........................................... (1)
Also ∵ sum of their squares is 189
∴ a²( 1/r² + 1 + r² ) = 189 ........................................... (2)
Squaring eq(1), we get :
... a² [ ( 1/r² + 1 + r² ) + 2( 1/r + 1 + r ) ] = 441
∴ a² ( 1/r² + 1 + r² ) + 2a· a( 1/r + 1 + r ) = 441
∴ 189 + 2a( 21 ) = 441 ∴ 2a(21) = 252
∴ a = 6 ........................................... (3)
∴ from (1) : 1/r + 1 + r = 21/a = 21/(6) =7/2
∴ 1/r + r = (7/2) - 1 = 5/2 ∴ (1/r) + r = (1/2) + 2 OR 2 + (1/2)
∴ r = 2 OR (1/2)............................ (4)
Putting (3) and (4) in (A) at the top,
we get the required numbers as : 3, 6, 12 ... OR ... 12, 6, 3 ......... Ans.