Gnpth
The sum of the terms of a geometric progression is 2047. Find the common ratio.
(1) The first and last terms of the series are 1 and 1024 respectively.
(2) Last but one term of the series is 512.
Summation of n terms of a G.P = sum = a\((\frac{ 1- r^n}{1-r})\)
Where a is the first term and n is the number of terms whose sum is being taken and r is the common ratio.
2047 = \(1 (\frac{ 1- r^n}{1-r})\)
Also nth term of a G.P.= \(ar^{(n-1)}\) where a is the first term and r is the common ratio.
So last term ( nth term) = 1024 = \(ar^{(n-1)}\)
We know a (first term) is 1
\(1024=r^{(n-1)}\)
\(2^{10}\) = 1024 here n-1= 10 so n=11
\(4^5\) = 1024 here n-1= 5 so n=6
\(32^2\) = 1024 here n-1= 2 so n=3
Out of these only n=11 will give sum of 2047 hence r= 2 is the required common ratio.
Hope this helps !
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Thank you.