Advertisements
Advertisements
Question
Find the sum of G.P. :
`1 - 1/3 + 1/3^2 - 1/3^3 + .........` to n terms.
Advertisements
Solution
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]`
RELATED QUESTIONS
Find, which of the following sequence from a G.P. :
8, 24, 72, 216, .............
Find the 9th term of the series :
1, 4, 16, 64, ...............
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.
If a, b and c are in G.P., prove that : log a, log b and log c are in A.P.
If a, b and c are in A.P, a, x, b are in G.P. whereas b, y and c are also in G.P.
Show that : x2, b2, y2 are in A.P.
Find the sum of G.P. :
`sqrt(3) + 1/sqrt(3) + 1/(3sqrt(3)) + ..........` to n terms.
Q 3.2
The first two terms of a G.P. are 125 and 25 respectively. Find the 5th and the 6th terms of the G.P.
Find a G.P. for which the sum of first two terms is – 4 and the fifth term is 4 times the third term.
