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`
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.
Q 6
Q 7
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
Q 7
Find the 5th term of the G.P. `5/2, 1, .........`
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.
