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प्रश्न
Find the sum of G.P. :
`1 - 1/3 + 1/3^2 - 1/3^3 + .........` to n terms.
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उत्तर
Given G.P. : `1 - 1/3 + 1/3^2 - 1/3^3 + .........` upto n terms
Here,
First term, a = 1
Common ratio, r = `(-1/3)/1 = -1/3` ...(∵ r < 1)
Number of terms to be added = n
∴ `S_n = (a(1 - r^n))/(1 - r)`
`=> S_n = (1(1 - (-1/3)^n))/(1 - (-1/3))`
= `(1(1 - (-1/3)^n))/(1 + 1/3)`
= `[1 - (-1/3)^n]/(4/3)`
= `3/4[1 - (-1/3)^n]`
संबंधित प्रश्न
Find, which of the following sequence from a G.P. :
`1/8, 1/24, 1/72, 1/216, ................`
Which term of the G.P.:
`-10, 5/sqrt(3), -5/6,....` is `-5/72`?
If the first and the third terms of a G.P. are 2 and 8 respectively, find its second term.
The product of 3rd and 8th terms of a G.P. is 243. If its 4th term is 3, find its 7th term.
The fourth term, the seventh term and the last term of a geometric progression are 10, 80 and 2560 respectively. Find its first term, common ratio and number of terms.
Q 7
If each term of a G.P. is raised to the power x, show that the resulting sequence is also a G.P.
Q 3.2
Q 8
The first term of a G.P. is –3 and the square of the second term is equal to its 4th term. Find its 7th term.
