Advertisements
Advertisements
प्रश्न
Find the sum of the following geometric series:
`sqrt7, sqrt21, 3sqrt7,...` to n terms
Advertisements
उत्तर
The given geometric series is:
`sqrt7, sqrt21, 3sqrt7,...` to n terms
Step 1: Identify the first term (a)
a = `sqrt7`
Step 2: Find the common ratio (r)
`r = (sqrt21)/(sqrt7) = sqrt3`
Check with next term:
`(3sqrt7)/(sqrt21) = sqrt3`
So the ratio is correct.
Step 3: Use the sum of n terms formula
For a geometric series:
`S_n = a(r^n - 1)/(r - 1)`
Substitute a = √7 and r = √3:
`S_n = sqrt7((sqrt3)^n - 1)/(sqrt3 - 1)`
संबंधित प्रश्न
Find the 20th and nthterms of the G.P. `5/2, 5/4 , 5/8,...`
Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015,…
Show that one of the following progression is a G.P. Also, find the common ratio in case:1/2, 1/3, 2/9, 4/27, ...
Find the 4th term from the end of the G.P.
In a GP the 3rd term is 24 and the 6th term is 192. Find the 10th term.
Find the sum of the following geometric progression:
(a2 − b2), (a − b), \[\left( \frac{a - b}{a + b} \right)\] to n terms;
Find the sum of the following geometric series:
\[\sqrt{2} + \frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}} + . . .\text { to 8 terms };\]
Evaluate the following:
\[\sum^{10}_{n = 2} 4^n\]
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.
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 \[\frac{1}{r^n}\].
A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of the terms occupying the odd places. Find the common ratio of the G.P.
If a, b, c, d are in G.P., prove that:
\[\frac{ab - cd}{b^2 - c^2} = \frac{a + c}{b}\]
If a, b, c are in G.P., then prove that:
If pth, qth, rth and sth terms of an A.P. be in G.P., then prove that p − q, q − r, r − s 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 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.
Find the geometric means of the following pairs of number:
a3b and ab3
The sum of two numbers is 6 times their geometric means, show that the numbers are in the ratio `(3+2sqrt2):(3-2sqrt2)`.
If logxa, ax/2 and logb x are in G.P., then write the value of x.
Write the product of n geometric means between two numbers a and b.
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 value of 91/3 . 91/9 . 91/27 ... upto inf, is
If a, b, c are in G.P. and x, y are AM's between a, b and b,c respectively, then
Given that x > 0, the sum \[\sum^\infty_{n = 1} \left( \frac{x}{x + 1} \right)^{n - 1}\] equals
Check whether the following sequence is G.P. If so, write tn.
`sqrt(5), 1/sqrt(5), 1/(5sqrt(5)), 1/(25sqrt(5))`, ...
For what values of x, the terms `4/3`, x, `4/27` are in G.P.?
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
For the following G.P.s, find Sn
3, 6, 12, 24, ...
For the following G.P.s, find Sn
0.7, 0.07, 0.007, .....
For a G.P. If t4 = 16, t9 = 512, find S10
The sum of an infinite G.P. is 5 and the sum of the squares of these terms is 15 find the G.P.
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 perimeters of all the squares
Select the correct answer from the given alternative.
The common ratio for the G.P. 0.12, 0.24, 0.48, is –
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 -
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
The sum of the first three terms of a G.P. is S and their product is 27. Then all such S lie in ______.
