Advertisements
Advertisements
प्रश्न
Find: `sum_("r" = 1)^10(3 xx 2^"r")`
उत्तर
`sum_("r" = 1)^10(3 xx 2^"r") = 3 sum_("r"=1)^10 2"r"`
= 3(2 + 22 + 23 + ... + 210)
Here, 2, 22, 23, ..., 210 are in G.P. with a = 2, r = 2
∴ `sum_("r" = 1)^10(3 xx 2^"r") = 3[(2(2^10 - 1))/(2 -1)]` ....... `[because "S"_"n" = ("a"("r"^"n" -1))/("r" -1)]`
= 6(1024 – 1)
= 6(1023)
= 6138
APPEARS IN
संबंधित प्रश्न
For what values of x, the numbers `-2/7, x, -7/2` are in G.P?
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.
A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.
Show that one of the following progression is a G.P. Also, find the common ratio in case:
4, −2, 1, −1/2, ...
Find :
the 8th term of the G.P. 0.3, 0.06, 0.012, ...
Find :
nth term of the G.P.
\[\sqrt{3}, \frac{1}{\sqrt{3}}, \frac{1}{3\sqrt{3}}, . . .\]
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.
The product of three numbers in G.P. is 125 and the sum of their products taken in pairs is \[87\frac{1}{2}\] . Find them.
Find the sum of the following geometric series:
`3/5 + 4/5^2 + 3/5^3 + 4/5^4 + ....` to 2n terms;
Find the sum of the following geometric series:
\[\sqrt{7}, \sqrt{21}, 3\sqrt{7}, . . .\text { to n terms }\]
Evaluate the following:
\[\sum^n_{k = 1} ( 2^k + 3^{k - 1} )\]
The 4th and 7th terms of a G.P. are \[\frac{1}{27} \text { and } \frac{1}{729}\] respectively. Find the sum of n terms of the G.P.
If S1, S2, ..., Sn are the sums of n terms of n G.P.'s whose first term is 1 in each and common ratios are 1, 2, 3, ..., n respectively, then prove that S1 + S2 + 2S3 + 3S4 + ... (n − 1) Sn = 1n + 2n + 3n + ... + nn.
Let an be the nth term of the G.P. of positive numbers.
Let \[\sum^{100}_{n = 1} a_{2n} = \alpha \text { and } \sum^{100}_{n = 1} a_{2n - 1} = \beta,\] such that α ≠ β. Prove that the common ratio of the G.P. is α/β.
Express the recurring decimal 0.125125125 ... as a rational number.
Find the rational numbers having the following decimal expansion:
\[0 .\overline {231 }\]
Find the rational numbers having the following decimal expansion:
\[0 . 6\overline8\]
If a, b, c are in G.P., prove that:
\[\frac{(a + b + c )^2}{a^2 + b^2 + c^2} = \frac{a + b + c}{a - b + c}\]
If a, b, c, d are in G.P., prove that:
(b + c) (b + d) = (c + a) (c + d)
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, d are in G.P., prove that:
(a2 − b2), (b2 − c2), (c2 − d2) are in G.P.
If a, b, c, d are in G.P., prove that:
\[\frac{1}{a^2 + b^2}, \frac{1}{b^2 - c^2}, \frac{1}{c^2 + d^2} \text { are in G . P } .\]
If a, b, c are in A.P., b,c,d are in G.P. and \[\frac{1}{c}, \frac{1}{d}, \frac{1}{e}\] are in A.P., prove that a, c,e are in G.P.
If A1, A2 be two AM's and G1, G2 be two GM's between a and b, then find the value of \[\frac{A_1 + A_2}{G_1 G_2}\]
If in an infinite G.P., first term is equal to 10 times the sum of all successive terms, then its common ratio is
The nth term of a G.P. is 128 and the sum of its n terms is 225. If its common ratio is 2, then its first term is
If a, b, c are in G.P. and x, y are AM's between a, b and b,c respectively, then
Check whether the following sequence is G.P. If so, write tn.
3, 4, 5, 6, …
Check whether the following sequence is G.P. If so, write tn.
7, 14, 21, 28, …
For the following G.P.s, find Sn.
`sqrt(5)`, −5, `5sqrt(5)`, −25, ...
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.
Determine whether the sum to infinity of the following G.P.s exist, if exists find them:
`-3, 1, (-1)/3, 1/9, ...`
Find : `sum_("r" = 1)^oo 4(0.5)^"r"`
Find `sum_("r" = 0)^oo (-8)(-1/2)^"r"`
Answer the following:
For a G.P. a = `4/3` and t7 = `243/1024`, find the value of r
Answer the following:
Find the nth term of the sequence 0.6, 0.66, 0.666, 0.6666, ...
Let S be the sum, P be the product and R be the sum of the reciprocals of 3 terms of a G.P. Then P2 R3 : S3 is equal to ______.