Advertisements
Advertisements
प्रश्न
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
Advertisements
उत्तर
Let the first term of the G.P. be a and its common ratio be R.
Then,
pth term = a `=>` ARp – 1 = a
qth term = b `=>` ARq – 1 = b
rth term = c `=>` ARr – 1 = c
Now,
aq – r × br – p × cp – q = (ARp – 1)q – r × (ARq – 1)r – p × (ARr – 1)p – q
= `A^(q - r) . R^((p - 1)(q - r)) xx A^(r - p) . R^((q - 1)(r - p)) xx A^(p - q) . R^((r - 1)(p - q))`
= `A^(q - r + r - p + p - q) xx R^((p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q))`
= A0 × R0
= 1
Taking log on both the sides, we get
log (aq – r × br – p × cp – q) = log 1
`=>` (q – r) log a + (r – p) log b + (p – q) log c = 0 ...(proved)
संबंधित प्रश्न
Which term of the G.P.:
`-10, 5/sqrt(3), -5/6,....` is `-5/72`?
The fifth term of a G.P. is 81 and its second term is 24. Find the geometric progression.
Fourth and seventh terms of a G.P. are `1/18` and `-1/486` respectively. Find the G.P.
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`
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.
Find the sum of G.P. : 3, 6, 12, .........., 1536.
The first two terms of a G.P. are 125 and 25 respectively. Find the 5th and the 6th terms of the G.P.
Find a G.P. for which the sum of first two terms is – 4 and the fifth term is 4 times the third term.
