Advertisements
Advertisements
प्रश्न
The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.
उत्तर
Let the first term of the geometric progression = a
Common and ratio = r
5th term = ar5–1 = ar4 = p
8th term = ar8–1 = ar7 = q
11th term = ar11–1= ar10 = s
Left side = q2 = (ar7)2
= a2 × r14
Right side = ps = ar4 ar10
= a2 × r14
Hence, q2 = ps
APPEARS IN
संबंधित प्रश्न
Find the 20th and nthterms of the G.P. `5/2, 5/4 , 5/8,...`
For what values of x, the numbers `-2/7, x, -7/2` are in G.P?
Find the sum to indicated number of terms of the geometric progressions `sqrt7, sqrt21,3sqrt7`...n terms.
The sum of first three terms of a G.P. is `39/10` and their product is 1. Find the common ratio and the terms.
Given a G.P. with a = 729 and 7th term 64, determine S7.
If a, b, c and d are in G.P. show that (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2 .
Find the value of n so that `(a^(n+1) + b^(n+1))/(a^n + b^n)` may be the geometric mean between a and b.
Show that one of the following progression is a G.P. Also, find the common ratio in case:
\[a, \frac{3 a^2}{4}, \frac{9 a^3}{16}, . . .\]
Which term of the G.P. :
\[\frac{1}{3}, \frac{1}{9}, \frac{1}{27} . . \text { . is } \frac{1}{19683} ?\]
If the G.P.'s 5, 10, 20, ... and 1280, 640, 320, ... have their nth terms equal, find the value of n.
Find the sum of the following geometric series:
(x +y) + (x2 + xy + y2) + (x3 + x2y + xy2 + y3) + ... to n terms;
Evaluate the following:
\[\sum^n_{k = 1} ( 2^k + 3^{k - 1} )\]
Find the sum of the following series:
9 + 99 + 999 + ... to n terms;
How many terms of the sequence \[\sqrt{3}, 3, 3\sqrt{3},\] ... must be taken to make the sum \[39 + 13\sqrt{3}\] ?
The common ratio of a G.P. is 3 and the last term is 486. If the sum of these terms be 728, find the first term.
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, S3 be respectively the sums of n, 2n, 3n terms of a G.P., then prove that \[S_1^2 + S_2^2\] = S1 (S2 + S3).
Find the sum of the following series to infinity:
10 − 9 + 8.1 − 7.29 + ... ∞
Find the rational numbers having the following decimal expansion:
\[0 . 6\overline8\]
Find an infinite G.P. whose first term is 1 and each term is the sum of all the terms which follow it.
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 are in G.P., prove that:
\[a^2 b^2 c^2 \left( \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} \right) = a^3 + b^3 + c^3\]
If a, b, c are three distinct real numbers in G.P. and a + b + c = xb, then prove that either x< −1 or x > 3.
If pth, qth and rth terms of an A.P. and G.P. are both a, b and c respectively, show that \[a^{b - c} b^{c - a} c^{a - b} = 1\]
Find the geometric means of the following pairs of number:
2 and 8
If logxa, ax/2 and logb x are in G.P., then write the value of x.
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
The product (32), (32)1/6 (32)1/36 ... to ∞ is equal to
For what values of x, the terms `4/3`, x, `4/27` are in G.P.?
For the following G.P.s, find Sn
3, 6, 12, 24, ...
For a G.P. if a = 2, r = 3, Sn = 242 find n
For a G.P. If t4 = 16, t9 = 512, find S10
The midpoints of the sides of a square of side 1 are joined to form a new square. This procedure is repeated indefinitely. Find the sum of the areas of all the squares
Select the correct answer from the given alternative.
If for a G.P. `"t"_6/"t"_3 = 1458/54` then r = ?
Answer the following:
Find `sum_("r" = 1)^"n" (2/3)^"r"`
Answer the following:
If pth, qth and rth terms of a G.P. are x, y, z respectively. Find the value of xq–r .yr–p .zp–q
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 the pth and qth terms of a G.P. are q and p respectively, show that its (p + q)th term is `(q^p/p^q)^(1/(p - q))`