# 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.

2
by jahanavialekya1

2015-03-22T17:01:28+05:30

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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
I think 16-9=7 but not 5
2015-03-23T00:28:05+05:30

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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), ....

Sir I think b+5=√601 /2+5=√601+1/2
sorry it was [√601+10]/2
i believe my answer is correct.
Ok Sir