Advertisements
Advertisements
प्रश्न
If each term of a G.P. is raised to the power x, show that the resulting sequence is also a G.P.
Advertisements
उत्तर
Let a1, a2, a3, ................., an, .......... be a G.P. with common ratio r.
`=> (a_(n + 1))/a_n = r` for all n ∈ N
If each term of a G.P. is raised to the power x, we get the sequence `a_1^x, a_2^x, a_3^x, ............, a_n^x,.........`
Now, `(a_(n + 1))^x/(a_n)^x = ((a_(n + 1))/a_n)^x = r^x` for all n ∈ N
Hence, `a_1^x, a_2^x, a_3^x, ............, a_n^x,.........` is also a G.P.
APPEARS IN
संबंधित प्रश्न
The product of 3rd and 8th terms of a G.P. is 243. If its 4th term is 3, find its 7th term.
Find the geometric progression with 4th term = 54 and 7th term = 1458.
Q 7
If a, b and c are in G.P., prove that : log a, log b and log c are in A.P.
Find the sum of G.P. :
`1 - 1/2 + 1/4 - 1/8 + ..........` to 9 terms.
Find the sum of G.P. :
`(x + y)/(x - y) + 1 + (x - y)/(x + y) + ..........` upto n terms.
How many terms of the geometric progression 1 + 4 + 16 + 64 + …….. must be added to get sum equal to 5461?
The sum of three numbers in G.P. is `39/10` and their product is 1. Find the numbers.
The first term of a G.P. is –3 and the square of the second term is equal to its 4th term. Find its 7th term.
Find a G.P. for which the sum of first two terms is – 4 and the fifth term is 4 times the third term.
