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
APPEARS IN
संबंधित प्रश्न
Find, which of the following sequence from a G.P. :
`1/8, 1/24, 1/72, 1/216, ................`
Which term of the G.P.:
`-10, 5/sqrt(3), -5/6,....` is `-5/72`?
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.
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. :
`sqrt(3) + 1/sqrt(3) + 1/(3sqrt(3)) + ..........` to n terms.
Q 2
Q 3.1
Q 3.2
Q 3.3
