Advertisements
Advertisements
Question
Answer the following:
Find `sum_("r" = 1)^"n" (2/3)^"r"`
Advertisements
Solution
`sum_("r" = 1)^"n" (2/3)^"r" = 2/3 + (2/3)^2 + (2/3)^3 + ... + (2/3)^"n"`
The terms `2/3, (2/3)^2, (2/3)^3` are in G.P.
∴ a = `2/3`, r = `2/3`
∴ `sum_("r" = 1)^"n" (2/3)^"r" = (2/3[1 - (2/3)^"n"])/(1 - 2/3)`
∴ `sum_("r" = 1)^"n" (2/3)^"r" = 2[1 - (2/3)^"n"]`
APPEARS IN
RELATED QUESTIONS
Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is `1/r^n`.
If f is a function satisfying f (x +y) = f(x) f(y) for all x, y ∈ N such that f(1) = 3 and `sum_(x = 1)^n` f(x) = 120, find the value of n.
The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.
Show that one of the following progression is a G.P. Also, find the common ratio in case:
\[a, \frac{3 a^2}{4}, \frac{9 a^3}{16}, . . .\]
Find:
the ninth term of the G.P. 1, 4, 16, 64, ...
Find:
the 10th term of the G.P.
\[- \frac{3}{4}, \frac{1}{2}, - \frac{1}{3}, \frac{2}{9}, . . .\]
Which term of the progression 0.004, 0.02, 0.1, ... is 12.5?
Which term of the progression 18, −12, 8, ... is \[\frac{512}{729}\] ?
The seventh term of a G.P. is 8 times the fourth term and 5th term is 48. Find the G.P.
Find the sum of the following geometric series:
\[\sqrt{2} + \frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}} + . . .\text { to 8 terms };\]
Find the sum of the following geometric series:
\[\frac{2}{9} - \frac{1}{3} + \frac{1}{2} - \frac{3}{4} + . . . \text { to 5 terms };\]
Find the sum of the following geometric series:
`sqrt7, sqrt21, 3sqrt7,...` to n terms
How many terms of the series 2 + 6 + 18 + ... must be taken to make the sum equal to 728?
If a and b are the roots of x2 − 3x + p = 0 and c, d are the roots x2 − 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (q − p) = 17 : 15.
If Sp denotes the sum of the series 1 + rp + r2p + ... to ∞ and sp the sum of the series 1 − rp + r2p − ... to ∞, prove that Sp + sp = 2 . S2p.
Find the rational number whose decimal expansion is `0.4bar23`.
Find the rational numbers having the following decimal expansion:
\[0 .\overline {231 }\]
If S denotes the sum of an infinite G.P. S1 denotes the sum of the squares of its terms, then prove that the first term and common ratio are respectively
\[\frac{2S S_1}{S^2 + S_1}\text { and } \frac{S^2 - S_1}{S^2 + S_1}\]
If a, b, c, d are in G.P., prove that:
(a + b + c + d)2 = (a + b)2 + 2 (b + c)2 + (c + d)2
If a, b, c are in G.P., prove that the following is also in G.P.:
a2, b2, c2
If a, b, c are in G.P., prove that the following is also in G.P.:
a2 + b2, ab + bc, b2 + c2
If a, b, c are in A.P. and a, b, d are in G.P., then prove that a, a − b, d − c are in G.P.
The sum of two numbers is 6 times their geometric means, show that the numbers are in the ratio `(3+2sqrt2):(3-2sqrt2)`.
If a, b, c are in G.P. and x, y are AM's between a, b and b,c respectively, then
If A be one A.M. and p, q be two G.M.'s between two numbers, then 2 A is equal to
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
The numbers 3, x, and x + 6 form are in G.P. Find nth term
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 t3 = 20 , t6 = 160 , find S7
For a sequence, if Sn = 2(3n –1), find the nth term, hence show that the sequence is a G.P.
Determine whether the sum to infinity of the following G.P.s exist, if exists find them:
`1/5, (-2)/5, 4/5, (-8)/5, 16/5, ...`
Answer the following:
If p, q, r, s are in G.P., show that (pn + qn), (qn + rn) , (rn + sn) are also in G.P.
If pth, qth, and rth terms of an A.P. and G.P. are both a, b and c respectively, show that ab–c . bc – a . ca – b = 1
If x, 2y, 3z are in A.P., where the distinct numbers x, y, z are in G.P. then the common ratio of the G.P. is ______.
Let `{a_n}_(n = 0)^∞` be a sequence such that a0 = a1 = 0 and an+2 = 2an+1 – an + 1 for all n ≥ 0. Then, `sum_(n = 2)^∞ a^n/7^n` is equal to ______.
