Advertisements
Advertisements
प्रश्न
Find the number of terms in the following G.P.
`1/3, 1/9, 1/27, ..., 1/2187`
उत्तर
`1/3, 1/9, 1/27, ..., 1/2187`
Hint: | |
3 | 729 |
3 | 243 |
3 | 81 |
3 | 27 |
3 | 9 |
3 | 3 |
1 |
Here a = `1/3`, r = `(1/9)/(1/3)`
= `1/9 xx 3/1`
= `1/3`
tn = arn–1
`1/3 xx (1/3)^("n" - 1)`
= `1/2817`
`(1/3)^("n" - 1) = 1/2817 xx 3`
= `1/729`
`(1/3)^("n" - 1) = 1/3^6`
= `(1/3)^6`
n – 1 = 6
n = 6 + 1 = 7
∴ No. of terms = 7
APPEARS IN
संबंधित प्रश्न
Identify the following sequence is in G.P.?
`1/3, 1/6, 1/12, ...`
Identify the following sequence is in G.P.?
1, −5, 25, −125, ...
Write the first three terms of the G.P. whose first term and the common ratio are given below
a = 6, r = 3
Write the first three terms of the G.P. whose first term and the common ratio are given below
a = `sqrt(2)`, r = `sqrt(2)`
Write the first three terms of the G.P. whose first term and the common ratio are given below
a = 1000, r = `2/5`
Find x so that x + 6, x + 12 and x + 15 are consecutive terms of a Geometric Progression
Find the 10th term of a G.P. whose 8th term is 768 and the common ratio is 2
If a, b, c is in A.P. then show that 3a, 3b, 3c is in G.P.
If a, b, c is three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that xb–c × yc–a × za–b = 1
The sum of the first 10 terms of the series 9 + 99 + 999 + ..., is ______.