Advertisements
Advertisements
Question
For a G.P., if a = 2, r = `-2/3`, find S6.
Solution
a = 2, r = `-2/3`
Sn = `("a"(1 - "r"^"n"))/(1 - "r")`, for r < 1
∴ S6 = `(2[1 - (-2/3)^6])/(1 - (-2/3)`
= `(2[1 - (-2/3)^6])/(5/3)`
= `6/5[(729 - 64)/3^6]`
= `6/5[665/729]`
∴ S6 = `266/243`.
APPEARS IN
RELATED QUESTIONS
For the following G.P.'s, find Sn: 3, 6, 12, 24, ...
For a G.P., if S5 = 1023, r = 4, find a.
For a G.P., if a = 2, r = 3, Sn = 242, find n.
For a G.P., if the sum of the first 3 terms is 125 and the sum of the next 3 terms is 27, find the value of r.
For a G.P., if t3 = 20, t6 = 160, find S7.
For a G.P., if t4 = 16, t9 = 512, find S10.
Find the sum to n terms: 8 + 88 + 888 + 8888 + …
Find the sum to n term: 0.4 + 0.44 + 0.444 + …
Find the sum to n terms: 0.7 + 0.77 + 0.777 + ...
If S, P, R are the sum, product and sum of the reciprocals of n terms of a G.P. respectively, then verify that `("S"/"R")^"n" = "P"^2`.
If for a sequence, tn = `5^(n-3)/2^(n-3)`, show that the sequence is a G.P.
Find its first term and the common ratio.
If Sn, S2n, S3n are the sum of n, 2n, 3n terms of a G.P. respectively, then verify that Sn (S3n - S2n) = (S2n - Sn)2.
If Sn, S2n, S3n are the sum of n, 2n, 3n terms of a G.P. respectively, then verify that Sn (S3n - S2n) = (S2n - Sn)2.
If for a sequence, `t_n=5^(n-3)/2^(n-3)`, show that the sequence is a G.P.
Find its first term and the common ratio.
If Sn, S2n, S3n are the sum of n, 2n, 3n terms of a G.P. respectively, then verify that Sn (S3n − S2n) = (S2n − Sn)2.
If for a sequence, `"t"_"n" = (5^"n"-3)/(2^"n"-3)`, show that sequence is a G.P.
Find its first term and the common ratio.
If `S_n, S_(2n), S_(3n)` are the sum of n, 2n, 3n terms of a G.P. respectively, then verify that `S_n (S_(3n) - S_(2n)) = (S_(2n) - S_n)^2`.
If `S_n, S_(2n), S_(3n)` are the sum of n, 2n, 3n terms of a G.P. respectively, then verify that `S_n(S_(3n) - S_(2n)) = (S_(2n) - S_n)^2`.
