Advertisements
Advertisements
प्रश्न
The sum of three numbers in G.P. is 14. If the first two terms are each increased by 1 and the third term decreased by 1, the resulting numbers are in A.P. Find the numbers.
The sum of three consecutive numbers of a G.P. is 14. If 1 is added in first and second term each and 1 subtracted from third, the new numbers form an A.P. Find the numbers.
Advertisements
उत्तर
Let the numbers be a, ar and ar2.
\[\text { Sum }= 14 \]
\[ \Rightarrow a + ar + a r^2 = 14 \]
\[ \Rightarrow a(1 + r + r^2 ) = 14 . . . \left( i \right)\]
According to the question,a + 1, ar + 1 and ar2 − 1 are in A.P.
\[\therefore 2\left( ar + 1 \right) = a + 1 + a r^2 - 1\]
\[ \Rightarrow 2ar + 2 = a + a r^2 \]
\[ \Rightarrow 2ar + 2 = 14 - ar [\text { From }\left( i \right)]\]
\[ \Rightarrow 3ar = 12 \]
\[ \Rightarrow a = \frac{4}{r} . . . \left( ii \right)\]
\[\text { Putting } a = \frac{4}{r} \text { in }\left( i \right)\]
\[ \Rightarrow \frac{4}{r}(1 + r + r^2 ) = 14\]
\[ \Rightarrow 4 r^2 - 10r + 4 = 0 \]
\[ \Rightarrow 4 r^2 - 8r - 2r + 4 = 0 \]
\[ \Rightarrow \left( 4r - 2 \right)\left( r - 2 \right) = 0\]
\[ \Rightarrow r = \frac{1}{2}, 2\]
\[\text { Putting } r = \frac{1}{2}\text { in } \left( ii \right), \text { we get a } = 8 . \]
\[\text { So, the G . P . is 8, 4 and } 2 . \]
Similarly putting r = 2 in (ii), we get a = 2.
So, the G.P is 2, 4 and 8.
APPEARS IN
संबंधित प्रश्न
Find the 20th and nthterms of the G.P. `5/2, 5/4 , 5/8,...`
If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.
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, ...
Which term of the G.P. :
\[2, 2\sqrt{2}, 4, . . .\text { is }128 ?\]
Which term of the G.P.: `sqrt3, 3, 3sqrt3`, ... is 729?
The seventh term of a G.P. is 8 times the fourth term and 5th term is 48. Find the G.P.
If the G.P.'s 5, 10, 20, ... and 1280, 640, 320, ... have their nth terms equal, find the value of n.
The product of three numbers in G.P. is 216. If 2, 8, 6 be added to them, the results are in A.P. Find the numbers.
Find three numbers in G.P. whose product is 729 and the sum of their products in pairs is 819.
Find the sum of the following geometric series:
\[\frac{2}{9} - \frac{1}{3} + \frac{1}{2} - \frac{3}{4} + . . . \text { to 5 terms };\]
Find the sum of the following series:
0.5 + 0.55 + 0.555 + ... to n terms.
Find the sum :
\[\sum^{10}_{n = 1} \left[ \left( \frac{1}{2} \right)^{n - 1} + \left( \frac{1}{5} \right)^{n + 1} \right] .\]
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 sum of the following series to infinity:
`1/3+1/5^2 +1/3^3+1/5^4 + 1/3^5 + 1/56+ ...infty`
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
\[\frac{2S S_1}{S^2 + S_1}\text { and } \frac{S^2 - S_1}{S^2 + S_1}\]
Find k such that k + 9, k − 6 and 4 form three consecutive terms of a G.P.
The sum of three numbers a, b, c in A.P. is 18. If a and b are each increased by 4 and c is increased by 36, the new numbers form a G.P. Find a, b, c.
If a, b, c are in A.P. and a, b, d are in G.P., then prove that a, a − b, d − c are in G.P.
Insert 6 geometric means between 27 and \[\frac{1}{81}\] .
Find the geometric means of the following pairs of number:
2 and 8
The sum of an infinite G.P. is 4 and the sum of the cubes of its terms is 92. The common ratio of the original G.P. is
If A be one A.M. and p, q be two G.M.'s between two numbers, then 2 A is equal to
If p, q be two A.M.'s and G be one G.M. between two numbers, then G2 =
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
Check whether the following sequence is G.P. If so, write tn.
7, 14, 21, 28, …
For the G.P. if r = − 3 and t6 = 1701, find a.
Find four numbers in G.P. such that sum of the middle two numbers is `10/3` and their product is 1
The numbers x − 6, 2x and x2 are in G.P. Find 1st term
For a sequence, if Sn = 2(3n –1), find the nth term, hence show that the sequence is a G.P.
A ball is dropped from a height of 10m. It bounces to a height of 6m, then 3.6m and so on. Find the total distance travelled by the ball
Select the correct answer from the given alternative.
Sum to infinity of a G.P. 5, `-5/2, 5/4, -5/8, 5/16,...` is –
Select the correct answer from the given alternative.
Which of the following is not true, where A, G, H are the AM, GM, HM of a and b respectively. (a, b > 0)
Answer the following:
In a G.P., the fourth term is 48 and the eighth term is 768. Find the tenth term
Answer the following:
For a sequence , if tn = `(5^("n" - 2))/(7^("n" - 3))`, verify whether the sequence is a G.P. If it is a G.P., find its first term and the common ratio.
Answer the following:
Find the sum of infinite terms of `1 + 4/5 + 7/25 + 10/125 + 13/6225 + ...`
The sum of infinite number of terms of a decreasing G.P. is 4 and the sum of the terms to m squares of its terms to infinity is `16/3`, then the G.P. is ______.
