How many terms of the geometric progression 1 + 4 + 16 + 64 + …….. must be added to get sum equal to 5461 ?
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Let n terms be added.
Common ratio = 41\dfrac{4}{1}14 = 4.
⇒S=a(rn−1)(r−1)..........(As∣r∣>1)⇒5461=1[(4)n−1]4−1⇒4n−13=5461⇒4n−1=16383⇒4n=16384⇒4n=47⇒n=7.\Rightarrow S = \dfrac{a(r^n - 1)}{(r - 1)} ……….(As |r| \gt 1)\\[1em] \Rightarrow 5461 = \dfrac{1[(4)^n - 1]}{4 - 1} \\[1em] \Rightarrow \dfrac{4^n - 1}{3} = 5461 \\[1em] \Rightarrow 4^n - 1 = 16383 \\[1em] \Rightarrow 4^n = 16384 \\[1em] \Rightarrow 4^n = 4^7 \\[1em] \Rightarrow n = 7.⇒S=(r−1)a(rn−1)……….(As∣r∣>1)⇒5461=4−11[(4)n−1]⇒34n−1=5461⇒4n−1=16383⇒4n=16384⇒4n=47⇒n=7.
Hence, 7 terms must be added to get a sum of 5461.
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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
3+13+133+.......\sqrt{3} + \dfrac{1}{\sqrt{3}} + \dfrac{1}{3\sqrt{3}} + …….3+31+331+……. to n terms
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?
A boy spends ₹ 10 on first day, ₹ 20 on second day, ₹ 40 on third day and so on. Find how much in all, will he spend in 12 days?