Write the first three terms of the Geometric Progression whose first term and the
common ratio are respectively the length and breadth of the rectangle whose area
is 144 sq.mts and the length exceeds the breadth by 5 mts.
Answers
Answered by
0
The first three terms of the g.p. = a,ar, ar^2
The first term of the G.P. is the length and common ratio is the breadth of the rectangle
breadth = x
Length = (x + 5)
L * B = area of rectangle
x (x + 5) = 144
x^2 + 5x = 144
x^2 + 5x - 144 = 0
x^2 + 16x - 9x - 144 = 0
x(x + 16) -9(x + 16) = 0
(x + 16) (x - 9) = 0
x = -16 or x = 9
Length = 9 + 5 = 14
Breadth = 9
The first three terms of g.p. = a, ar, ar^2
a = 14
ar = 14*9 = 126
ar^2 = 14*9^2 = 14*81= 1134
The first three terms of the g.p. is 14, 126, 1134
AlekhyaSistu:
I think 16-9=7 but not 5
Answered by
3
rectangle: dimensions: l by b in meters.
Area = l b = 144
l = b + 5
=> (b + 5) b = 144
=> b² + 5 b - 144 = 0
=> b = [ -5 + - √(25 + 576) ] /2 = [-5 + √601 ] /2 as breadth is +ve.
length l = b+5 = [√601 + 5 ]/2
GP = l, l b, l b², ....
= (√601 + 5) / 2 , 144, 72(√601 - 5), ....
Area = l b = 144
l = b + 5
=> (b + 5) b = 144
=> b² + 5 b - 144 = 0
=> b = [ -5 + - √(25 + 576) ] /2 = [-5 + √601 ] /2 as breadth is +ve.
length l = b+5 = [√601 + 5 ]/2
GP = l, l b, l b², ....
= (√601 + 5) / 2 , 144, 72(√601 - 5), ....
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