Find the sum of G.P. :
x+yx−y+1+x−yx+y\dfrac{x + y}{x - y} + 1 + \dfrac{x - y}{x + y}x−yx+y+1+x+yx−y + …….. upto n terms
13 Likes
Common ratio (r) = 1x+yx−y=x−yx+y\dfrac{1}{\dfrac{x + y}{x - y}} = \dfrac{x - y}{x + y}x−yx+y1=x+yx−y.
S=a(1−rn)(1−r)..........(As∣r∣<1)=x+yx−y[1−(x−yx+y)n]1−x−yx+y=(x+y)[1−(x−yx+y)n](x−y)x+y−x+yx+y=(x+y)2[1−(x−yx+y)n]2y(x−y)=(x+y)22y(x−y)[1−(x−yx+y)n].S = \dfrac{a(1 - r^n)}{(1 - r)} ……….(As |r| \lt 1) \\[1em] = \dfrac{\dfrac{x + y}{x - y}\Big[1 - \Big(\dfrac{x - y}{x + y}\Big)^n\Big]}{1 - \dfrac{x - y}{x + y}} \\[1em] = \dfrac{(x + y)\Big[1 - \Big(\dfrac{x - y}{x + y}\Big)^n\Big]}{(x - y)\dfrac{x + y - x + y}{x + y}} \\[1em] = \dfrac{(x + y)^2\Big[1 - \Big(\dfrac{x - y}{x + y}\Big)^n\Big]}{2y(x - y)} \\[1em] = \dfrac{(x + y)^2}{2y(x - y)}\Big[1 - \Big(\dfrac{x - y}{x + y}\Big)^n\Big].S=(1−r)a(1−rn)……….(As∣r∣<1)=1−x+yx−yx−yx+y[1−(x+yx−y)n]=(x−y)x+yx+y−x+y(x+y)[1−(x+yx−y)n]=2y(x−y)(x+y)2[1−(x+yx−y)n]=2y(x−y)(x+y)2[1−(x+yx−y)n].
Hence, sum = (x+y)22y(x−y)[1−(x−yx+y)n].\dfrac{(x + y)^2}{2y(x - y)}\Big[1 - \Big(\dfrac{x - y}{x + y}\Big)^n\Big].2y(x−y)(x+y)2[1−(x+yx−y)n].
Answered By
8 Likes
Which term of the G.P. 2,22,4,........ is 1282?2, 2\sqrt{2}, 4, …….. \text{ is } 128\sqrt{2} ?2,22,4,…….. is 1282?
Find the 8th term of a G.P., if its common ratio is 2 and 10th term is 768.
In a G.P., the 4th term is 48 and 7th term is 384. Find its 6th term.
−53,x and −35-\dfrac{5}{3}, x \text{ and } -\dfrac{3}{5}−35,x and −53 are three consecutive terms of a G.P. Find the value(s) of x.
