Advertisements
Advertisements
प्रश्न
Answer the following:
For a G.P. if t2 = 7, t4 = 1575 find a
Advertisements
उत्तर
Given, t2 = 7, t4 = 1575
tn = arn–1
∴ t2 = ar
∴ ar = 7
∴ r = `7/"a"` ...(i)
Also, t4 = ar3
∴ ar3 = 1575
∴ `"a" xx (7/"a")^3` = 1575 ...[From (i)]
∴ a2 = `7^3/1575`
∴ a2 = `49/225`
∴ a = `7/15`
APPEARS IN
संबंधित प्रश्न
Find the 20th and nthterms of the G.P. `5/2, 5/4 , 5/8,...`
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 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.
If a, b, c are in A.P,; b, c, d are in G.P and ` 1/c, 1/d,1/e` are in A.P. prove that a, c, e are in G.P.
Find :
the 12th term of the G.P.
\[\frac{1}{a^3 x^3}, ax, a^5 x^5 , . . .\]
Find :
nth term of the G.P.
\[\sqrt{3}, \frac{1}{\sqrt{3}}, \frac{1}{3\sqrt{3}}, . . .\]
Find three numbers in G.P. whose sum is 65 and whose product is 3375.
Find the sum of the following geometric progression:
1, −1/2, 1/4, −1/8, ... to 9 terms;
Find the sum of the following geometric series:
(x +y) + (x2 + xy + y2) + (x3 + x2y + xy2 + y3) + ... to n terms;
Find the sum of the following geometric series:
\[\frac{a}{1 + i} + \frac{a}{(1 + i )^2} + \frac{a}{(1 + i )^3} + . . . + \frac{a}{(1 + i )^n} .\]
How many terms of the sequence \[\sqrt{3}, 3, 3\sqrt{3},\] ... must be taken to make the sum \[39 + 13\sqrt{3}\] ?
The sum of n terms of the G.P. 3, 6, 12, ... is 381. Find the value of n.
The fifth term of a G.P. is 81 whereas its second term is 24. Find the series and sum of its first eight terms.
Find the sum of the following serie to infinity:
\[1 - \frac{1}{3} + \frac{1}{3^2} - \frac{1}{3^3} + \frac{1}{3^4} + . . . \infty\]
Find the rational numbers having the following decimal expansion:
\[0 . \overline3\]
Find the rational numbers having the following decimal expansion:
\[0 . 6\overline8\]
One side of an equilateral triangle is 18 cm. The mid-points of its sides are joined to form another triangle whose mid-points, in turn, are joined to form still another triangle. The process is continued indefinitely. Find the sum of the (i) perimeters of all the triangles. (ii) areas of all triangles.
If a, b, c are in G.P., prove that \[\frac{1}{\log_a m}, \frac{1}{\log_b m}, \frac{1}{\log_c m}\] are in A.P.
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), (b − c), (c − a) are in G.P., then prove that (a + b + c)2 = 3 (ab + bc + ca)
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 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.
If pth, qth and rth terms of an A.P. and G.P. are both a, b and c respectively, show that \[a^{b - c} b^{c - a} c^{a - b} = 1\]
Insert 6 geometric means between 27 and \[\frac{1}{81}\] .
Find the geometric means of the following pairs of number:
−8 and −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 value of 91/3 . 91/9 . 91/27 ... upto inf, is
For the G.P. if a = `7/243`, r = 3 find t6.
For the following G.P.s, find Sn.
`sqrt(5)`, −5, `5sqrt(5)`, −25, ...
Find the sum to n terms of the sequence.
0.2, 0.02, 0.002, ...
Select the correct answer from the given alternative.
The tenth term of the geometric sequence `1/4, (-1)/2, 1, -2,` ... is –
Answer the following:
For a sequence Sn = 4(7n – 1) verify that the sequence is a G.P.
Answer the following:
Find the sum of infinite terms of `1 + 4/5 + 7/25 + 10/125 + 13/6225 + ...`
The sum of the first three terms of a G.P. is S and their product is 27. Then all such S lie in ______.
