Advertisements
Advertisements
Question
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, ...
Advertisements
Solution
We have,
\[ a_1 = \frac{1}{2} , a_2 = \frac{1}{3}, a_3 = \frac{2}{9}, a_4 = \frac{4}{27}\]
\[\text { Now }, \frac{a_2}{a_1} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}, \frac{a_3}{a_2} = \frac{\frac{2}{9}}{\frac{1}{3}} = \frac{2}{3}, \frac{a_4}{a_3} = \frac{\frac{4}{27}}{\frac{2}{9}} = \frac{2}{3}\]
\[ \therefore \frac{a_2}{a_1} = \frac{a_3}{a_2} = \frac{a_4}{a_3} = \frac{2}{3}\]
\[\text { Thus, } a_1 , a_2 , a_3 \text { and } a_4 \text { are in G . P . , where the first term is} \frac{1}{2} \text { and the common ratio is } \frac{2}{3} .\]
RELATED QUESTIONS
The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio `(3 + 2sqrt2) ":" (3 - 2sqrt2)`.
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.
Find :
nth term of the G.P.
\[\sqrt{3}, \frac{1}{\sqrt{3}}, \frac{1}{3\sqrt{3}}, . . .\]
Which term of the G.P.: `sqrt3, 3, 3sqrt3`, ... is 729?
Which term of the G.P. :
\[\frac{1}{3}, \frac{1}{9}, \frac{1}{27} . . \text { . is } \frac{1}{19683} ?\]
Find the 4th term from the end of the G.P.
\[\frac{1}{2}, \frac{1}{6}, \frac{1}{18}, \frac{1}{54}, . . . , \frac{1}{4374}\]
The seventh term of a G.P. is 8 times the fourth term and 5th term is 48. Find the G.P.
The sum of first three terms of a G.P. is \[\frac{39}{10}\] and their product is 1. Find the common ratio and the terms.
Find the sum of the following geometric progression:
1, 3, 9, 27, ... to 8 terms;
Find the sum of the following geometric series:
0.15 + 0.015 + 0.0015 + ... to 8 terms;
Find the sum of the following geometric series:
\[\frac{a}{1 + i} + \frac{a}{(1 + i )^2} + \frac{a}{(1 + i )^3} + . . . + \frac{a}{(1 + i )^n} .\]
Evaluate the following:
\[\sum^{11}_{n = 1} (2 + 3^n )\]
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.
How many terms of the G.P. `3, 3/2, 3/4` ..... are needed to give the sum `3069/512`?
Prove that: (91/3 . 91/9 . 91/27 ... ∞) = 3.
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.
Find an infinite G.P. whose first term is 1 and each term is the sum of all the terms which follow it.
If a, b, c are in G.P., prove that:
a (b2 + c2) = c (a2 + b2)
If a, b, c, d are in G.P., prove that:
\[\frac{ab - cd}{b^2 - c^2} = \frac{a + c}{b}\]
Insert 5 geometric means between 16 and \[\frac{1}{4}\] .
If x is positive, the sum to infinity of the series \[\frac{1}{1 + x} - \frac{1 - x}{(1 + x )^2} + \frac{(1 - x )^2}{(1 + x )^3} - \frac{(1 - x )^3}{(1 + x )^4} + . . . . . . is\]
In a G.P. of even number of terms, the sum of all terms is five times the sum of the odd terms. The common ratio of the G.P. is
Mosquitoes are growing at a rate of 10% a year. If there were 200 mosquitoes in the beginning. Write down the number of mosquitoes after 3 years.
Mosquitoes are growing at a rate of 10% a year. If there were 200 mosquitoes in the beginning. Write down the number of mosquitoes after n years.
For a G.P. a = 2, r = `-2/3`, find S6
Find the sum to n terms of the sequence.
0.2, 0.02, 0.002, ...
Determine whether the sum to infinity of the following G.P.s exist, if exists find them:
`2, 4/3, 8/9, 16/27, ...`
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, ...`
If the common ratio of a G.P. is `2/3` and sum to infinity is 12. Find the first term
Find : `sum_("n" = 1)^oo 0.4^"n"`
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 -
Answer the following:
Find `sum_("r" = 1)^"n" (2/3)^"r"`
Answer the following:
If a, b, c are in G.P. and ax2 + 2bx + c = 0 and px2 + 2qx + r = 0 have common roots then verify that pb2 – 2qba + ra2 = 0
In a G.P. of even number of terms, the sum of all terms is 5 times the sum of the odd terms. The common ratio of the G.P. is ______.
For a, b, c to be in G.P. the value of `(a - b)/(b - c)` is equal to ______.
The third term of a G.P. is 4, the product of the first five terms is ______.
The sum or difference of two G.P.s, is again a G.P.
