Advertisements
Advertisements
प्रश्न
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?
उत्तर
a. Given, G.P. is 15, 30, 60, 120....
Here, a = 15
Common ratio (r) = `30/15` = 2
Then an = arn – 1
= 15(2)n – 1
b. Sum of n terms,
`S_n = (a(r^n - 1))/(r - 1)` ...(∵ r > 1)
`\implies 945 = 15((2^n - 1))/(2 - 1)`
`\implies 945/15 = 2^n - 1`
`\implies` 63 = 2n – 1
`\implies` 2n = 64
`\implies` 2n = 26
∴ n = 6
Hence, number of terms needed are 6.
APPEARS IN
संबंधित प्रश्न
Find the 8th term of the sequence:
`3/4, 1 1/2, 3, ..............`
Find the 10th term of the G.P. :
`12, 4, 1 1/3, ................`
Find the seventh term of the G.P. :
`sqrt(3) + 1, 1, (sqrt(3) - 1)/2, .........`
Find the next three terms of the series :
`2/27, 2/9, 2/3, .............`
A geometric progression has common ratio = 3 and last term = 486. If the sum of its terms is 728; find its first term.
Q 10.2
Find how many terms of G.P. `2/9 - 1/3 + 1/2` ......... must be added to get the sum equal to `55/72`?
Q 3.2
Q 3.4
Q 4