Advertisements
Advertisements
प्रश्न
The fifth, eight and eleventh terms of a geometric progression are p, q and r respectively. Show that : q2 = pr.
Advertisements
उत्तर
Let the first term of the G.P. be a and its common ratio be r.
5th term = t5 = p
`=>` ar4 = p
8th term = t8 = q
`=>` ar7 = q
11th term = t11 = r
`=>` ar10 = r
Now,
pr = ar4 × ar10
= a2 × r14
= (a × r7)2
= q2
`=>` q2 = pr
संबंधित प्रश्न
Find the 9th term of the series :
1, 4, 16, 64, ...............
For the G.P. `1/27, 1/9, 1/3, ........., 81`; find the product of fourth term from the beginning and the fourth term from the end.
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.
If a, b, c are in G.P. and a, x, b, y, c are in A.P., prove that `a/x + c/y = 2`
Q 6
If a, b and c are in A.P. and also in G.P., show that : a = b = c.
Find the sum of G.P. :
`1 - 1/3 + 1/3^2 - 1/3^3 + .........` to n terms.
Find the sum of G.P. :
`sqrt(3) + 1/sqrt(3) + 1/(3sqrt(3)) + ..........` to n terms.
Find the geometric mean between 2a and 8a3
The first two terms of a G.P. are 125 and 25 respectively. Find the 5th and the 6th terms of the G.P.
