Advertisements
Advertisements
Question
Find the seventh term of the G.P. :
`sqrt(3) + 1, 1, (sqrt(3) - 1)/2, .........`
Solution
Given G.P. : `sqrt(3) + 1, 1, (sqrt(3) - 1)/2, .........`
Here,
First term, a = `sqrt(3) + 1`
Common ratio, r = `1/(sqrt(3) + 1)`
Now, Tn = arn – 1
`\implies` T7 = `(sqrt(3) + 1) xx (1/(sqrt(3) + 1))^(7 - 1)`
= `(sqrt(3) + 1) xx (1/(sqrt(3) + 1))^6`
= `(sqrt(3) + 1)/1 xx 1/(sqrt(3) + 1)^6`
= `1/(sqrt(3) + 1)^5`
= `1/(sqrt(3) + 1)^5 xx (sqrt(3) - 1)^5/(sqrt(3) - 1)^5`
= `(sqrt(3) - 1)^5/[(sqrt(3) + 1)(sqrt(3) - 1)]^5`
= `(sqrt(3) - 1)^5/(3 - 1)^5`
= `(sqrt(3) - 1)^5/(2)^5`
= `(sqrt(3) - 1)^5/32`
APPEARS IN
RELATED QUESTIONS
Find the seventh term of the G.P. :
`1, sqrt(3), 3, 3sqrt(3), ............`
Find the nth term of the series :
1, 2, 4, 8, ...............
Find the next three terms of the series :
`2/27, 2/9, 2/3, .............`
Q 10.4
If the sum of 1 + 2 + 22 + ....... + 2n – 1 is 255, find the value of n.
Q 1.2
Q 3.1
Q 10
The first and last term of a Geometrical Progression (G.P.) and 3 and 96 respectively. If the common ratio is 2, find:
(i) ‘n’ the number of terms of the G.P.
(ii) Sum of the n terms.
15, 30, 60, 120.... are in G.P. (Geometric Progression):
- Find the nth term of this G.P. in terms of n.
- How many terms of the above G.P. will give the sum 945?