Find the sum of G.P. :
3+13+133+.......\sqrt{3} + \dfrac{1}{\sqrt{3}} + \dfrac{1}{3\sqrt{3}} + …….3+31+331+……. to n terms
2 Likes
Common ratio (r) = 133=13.\dfrac{\dfrac{1}{\sqrt{3}}}{\sqrt{3}} = \dfrac{1}{3}.331=31.
S=a(1−rn)(1−r)..........(As∣r∣<1)=3×[1−(13)n]1−13=3×[1−(13)n]23=33×[1−(13)n]2=332(1−13n)S = \dfrac{a(1 - r^n)}{(1 - r)} ……….(As |r| \lt 1) \\[1em] = \dfrac{\sqrt{3} \times \Big[1 - \Big(\dfrac{1}{3}\Big)^n\Big]}{1 - \dfrac{1}{3}} \\[1em] = \dfrac{\sqrt{3} \times \Big[1 - \Big(\dfrac{1}{3}\Big)^n\Big]}{\dfrac{2}{3}} \\[1em] = \dfrac{3\sqrt{3} \times \Big[1 - \Big(\dfrac{1}{3}\Big)^n\Big]}{2} \\[1em] = \dfrac{3\sqrt{3}}{2}\Big(1 - \dfrac{1}{3^n}\Big)S=(1−r)a(1−rn)……….(As∣r∣<1)=1−313×[1−(31)n]=323×[1−(31)n]=233×[1−(31)n]=233(1−3n1)
Hence, sum = 332(1−13n)\dfrac{3\sqrt{3}}{2}\Big(1 - \dfrac{1}{3^n}\Big)233(1−3n1).
Answered By
3 Likes
1−13+132−1331 - \dfrac{1}{3} + \dfrac{1}{3^2} - \dfrac{1}{3^3}1−31+321−331 + …… to n terms
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
How many terms of the geometric progression 1 + 4 + 16 + 64 + …….. must be added to get sum equal to 5461 ?
The first term of a G.P. is 27. If the 8th term be 181\dfrac{1}{81}811, what will be the sum of 10 terms?