Advertisements
Advertisements
प्रश्न
Find the seventh term from the end of the series :
`sqrt(2), 2, 2sqrt(2), ........., 32.`
उत्तर
Given series: `sqrt(2), 2, 2sqrt(2), ........., 32.`
Now, `2/sqrt(2) = (2sqrt(2))/2 = sqrt(2)`
So, the given series is a G.P. with common ratio, r = `sqrt(2)`
Here, last term, l = 32
∴ 7th term from an end = `1/r^6`
= `32/(sqrt(2))^6`
= `32/8`
= 4
APPEARS IN
संबंधित प्रश्न
Find the 9th term of the series :
1, 4, 16, 64, ...............
If the first and the third terms of a G.P. are 2 and 8 respectively, find its second term.
For the G.P. `1/27, 1/9, 1/3, ........., 81`; find the product of fourth term from the beginning and the fourth term from the end.
Q 6
If a, b and c are in G.P., prove that : log a, log b and log c are in A.P.
Q 2
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`
Find the sum of G.P. :
`(x + y)/(x - y) + 1 + (x - y)/(x + y) + ..........` upto n terms.
Find the geometric mean between `4/9` and `9/4`
