Advertisements
Advertisements
Question
Find: `sum_("r" = 1)^10 5 xx 3^"r"`
Solution
`sum_("r" = 1)^10 5 xx 3^"r"`
= 5 x 3 + 5 x 32 + 5 x 33 + … + 5 x 310
= 5[3 + 32 + 33 + ... + 310]
Here, 3, 32, 33, ..., 310 are the terms in G.P. with
a = 3, r = 3
`sum_("r" = 1)^10 5 xx 3^"r" = 5*(3[3^10 - 1])/(3 - 1)`
= `15/2(3^10 - 1)`
APPEARS IN
RELATED QUESTIONS
Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.
If the pth , qth and rth terms of a G.P. are a, b and c, respectively. Prove that `a^(q - r) b^(r-p) c^(p-q) = 1`
Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio `(3 + 2sqrt2) ":" (3 - 2sqrt2)`.
Find :
nth term of the G.P.
\[\sqrt{3}, \frac{1}{\sqrt{3}}, \frac{1}{3\sqrt{3}}, . . .\]
Which term of the G.P. :
\[2, 2\sqrt{2}, 4, . . .\text { is }128 ?\]
Find the 4th term from the end of the G.P.
\[\frac{1}{2}, \frac{1}{6}, \frac{1}{18}, \frac{1}{54}, . . . , \frac{1}{4374}\]
The seventh term of a G.P. is 8 times the fourth term and 5th term is 48. Find the G.P.
If 5th, 8th and 11th terms of a G.P. are p. q and s respectively, prove that q2 = ps.
If \[\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}\] (x ≠ 0), then show that a, b, c and d are in G.P.
Find three numbers in G.P. whose sum is 38 and their product is 1728.
Find the sum of the following series:
0.6 + 0.66 + 0.666 + .... to n terms
Find the sum :
\[\sum^{10}_{n = 1} \left[ \left( \frac{1}{2} \right)^{n - 1} + \left( \frac{1}{5} \right)^{n + 1} \right] .\]
Find the sum of the following series to infinity:
10 − 9 + 8.1 − 7.29 + ... ∞
If xa = xb/2 zb/2 = zc, then prove that \[\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\] are in A.P.
If a, b, c are in A.P. and a, x, b and b, y, c are in G.P., show that x2, b2, y2 are in A.P.
The sum of two numbers is 6 times their geometric means, show that the numbers are in the ratio \[(3 + 2\sqrt{2}) : (3 - 2\sqrt{2})\] .
If S be the sum, P the product and R be the sum of the reciprocals of n terms of a GP, then P2 is equal to
The sum of an infinite G.P. is 4 and the sum of the cubes of its terms is 92. The common ratio of the original G.P. is
If second term of a G.P. is 2 and the sum of its infinite terms is 8, then its first term is
Let x be the A.M. and y, z be two G.M.s between two positive numbers. Then, \[\frac{y^3 + z^3}{xyz}\] is equal to
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
The fifth term of a G.P. is x, eighth term of a G.P. is y and eleventh term of a G.P. is z verify whether y2 = xz
The numbers 3, x, and x + 6 form are in G.P. Find x
For the following G.P.s, find Sn
3, 6, 12, 24, ...
For the following G.P.s, find Sn.
`sqrt(5)`, −5, `5sqrt(5)`, −25, ...
For a G.P. if a = 2, r = 3, Sn = 242 find n
For a sequence, if Sn = 2(3n –1), find the nth term, hence show that the sequence is a G.P.
Find : `sum_("r" = 1)^oo 4(0.5)^"r"`
Select the correct answer from the given alternative.
If for a G.P. `"t"_6/"t"_3 = 1458/54` then r = ?
Select the correct answer from the given alternative.
If common ratio of the G.P is 5, 5th term is 1875, the first term is -
Select the correct answer from the given alternative.
Which of the following is not true, where A, G, H are the AM, GM, HM of a and b respectively. (a, b > 0)
Answer the following:
For a G.P. if t2 = 7, t4 = 1575 find a
Answer the following:
Find k so that k – 1, k, k + 2 are consecutive terms of a G.P.
If a, b, c, d are in G.P., prove that a2 – b2, b2 – c2, c2 – d2 are also in G.P.
