Advertisements
Advertisements
Question
Which term of the G. P. 5, 25, 125, 625, … is 510?
Solution
Here, t1 = a = 5, r = `"t"_2/"t"_1 = 25/5` = 5, tn = 510
tn = arn – 1
∴ 510 = 5 × 5(n–1)
∴ 510 = `5^((1 + "n" - 1)`
∴ 510 = 5n
∴ n = 10
∴ 510 is the 10th term of the G.P.
APPEARS IN
RELATED QUESTIONS
The fifth term of a G. P. is x, eighth term of the G. P. is y and eleventh term of the G. P. is z. Verify whether y2 = xz.
For a sequence, if Sn = 2 (3n – 1), find the nth term, hence show that the sequence is a G.P.
In a G.P., the fourth term is 48 and the eighth term is 768. Find the tenth term.
Find 2 + 22 + 222 + 2222 + … upto n terms.
If for a sequence, `t_n = (5^(n- 3))/(2^(n - 3))`, show that the sequence is a G.P.
Find its first term and the common ratio.
For the G.P. if a = `2/3` t6 = 162, find r.
Express the following recurring decimals as a rational number.
`4.bar18`
For the G.P. if a = `2/3`, t6 = 162, find r.
Verify whether the following sequence is G.P. If so, find tn.
`sqrt5 , 1/sqrt5 , 1/(5sqrt5) , 1/(25sqrt5)`, ...
For the G.P. if a = `2/3`, t6 = 162, find r.
Verify whether the following sequences are G.P. If so, find tn.
`sqrt5, 1/sqrt5, 1/(5sqrt5), 1/(25sqrt5), ...`
For the G.P. if a = `2/3` , t6 = 162 , find r
For the G.P. If a = `2/3, t_6 = 162,` find r
For the G.P. if a = `2/3`, t6 = 162, find r.
