Advertisements
Advertisements
प्रश्न
The fifth, eight and eleventh terms of a geometric progression are p, q and r respectively. Show that : q2 = pr.
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. :
8, 24, 72, 216, .............
Find, which of the following sequence from a G.P. :
9, 12, 16, 24, ................
Find the geometric progression with 4th term = 54 and 7th term = 1458.
If for a G.P., pth, qth and rth terms are a, b and c respectively; prove that : (q – r) log a + (r – p) log b + (p – q) log c = 0
Q 5
Q 8
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.
Find the sum of G.P. :
`1 - 1/2 + 1/4 - 1/8 + ..........` to 9 terms.
Find the sum of G.P. :
`sqrt(3) + 1/sqrt(3) + 1/(3sqrt(3)) + ..........` to n terms.
Find the geometric mean between 14 and `7/32`