Advertisements
Advertisements
Question
If a, b and c are in A.P, a, x, b are in G.P. whereas b, y and c are also in G.P.
Show that : x2, b2, y2 are in A.P.
Advertisements
Solution
a, b and c are in A.P.
`=>` 2b = a + c
a, x and b are in G.P.
`=>` x2 = ab
b, y and c are in G.P.
`=>` y2 = bc
Now,
x2 + y2 = ab + bc
= b(a + c)
= b × 2b
= 2b2
`=>` x2, b2 and y2 are in A.P.
RELATED QUESTIONS
Find, which of the following sequence from a G.P. :
`1/8, 1/24, 1/72, 1/216, ................`
Find the 9th term of the series :
1, 4, 16, 64, ...............
Find the G.P. whose first term is 64 and next term is 32.
Find the geometric progression with 4th term = 54 and 7th term = 1458.
Second term of a geometric progression is 6 and its fifth term is 9 times of its third term. Find the geometric progression. Consider that each term of the G.P. is positive.
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 - 1/3 + 1/3^2 - 1/3^3 + .........` to n 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 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.
