Advertisements
Advertisements
प्रश्न
The sum of three numbers in G.P. is `39/10` and their product is 1. Find the numbers.
Advertisements
उत्तर
Let the number be `a/r`, a and ar.
`=> a/r xx a xx ar = 1`
`=>` a3 = 1
`=>` a = 1
Now, `a/r + a + ar = 39/10`
`=> 1/r + 1 + r = 39/10`
`=> (1 + r + r^2)/r = 39/10`
`=>` 10 + 10r + 10r2 = 39r
`=>` 10r2 – 29r + 10 = 0
`=>` 10r2 – 25r – 4r + 10 = 0
`=>` 5r(2r – 5) – 2(2r – 5) = 0
`=>` (2r – 5)(5r – 2) = 0
`=> r = 5/2` or `r = 2/5`
Thus, required terms are :
`a/r, a, ar = 1/(5/2), 1, 1 xx 5/2` or `1/(2/5), 1, 1 xx 2/5`
= `2/5, 1, 5/2` or `5/2, 1, 2/5`
APPEARS IN
संबंधित 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
Q 7
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 `1/x + 1/y = 2/b`
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`
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 + 3 + 9 + 27 + .......... to 12 terms.
Find the sum of G.P. :
0.3 + 0.03 + 0.003 + 0.0003 + ........... to 8 items.
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.
