Advertisements
Advertisements
प्रश्न
Find the sum of G.P. :
`1 - 1/3 + 1/3^2 - 1/3^3 + .........` to n terms.
Advertisements
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
Second term of a geometric progression is 6 and its fifth term is 9 times of its third term. Find the geometric progression. Consider that each term of the G.P. is positive.
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.
Find the third term from the end of the G.P.
`2/27, 2/9, 2/3, .........,162.`
If for a G.P., pth, qth and rth terms are a, b and c respectively; prove that : (q – r) log a + (r – p) log b + (p – q) log c = 0
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 6
Find the sum of G.P. :
`(x + y)/(x - y) + 1 + (x - y)/(x + y) + ..........` upto n terms.
Q 8
The first two terms of a G.P. are 125 and 25 respectively. Find the 5th and the 6th terms of the G.P.
