Advertisements
Advertisements
प्रश्न
For a G.P. If t4 = 16, t9 = 512, find S10
Advertisements
उत्तर
t4 = 16, t9 = 512
tn = arn–1
∴ t4 = ar4–1 = ar3
∴ ar3 = 16
∴ a = `16/"r"^3` ...(i)
Also, t9 = ar8
ar8 = 512
∴ `16/"r"^3 xx"r"^8` = 512
∴ r5 = 32
∴ r = 2
Substituting r = 2 in (i), we get
a `16/2^3`
= `16/8`
= 2
Now, Sn = `("a"("r"^"n"- 1))/("r" - 1)`, for r > 1
∴ S10 = `(2(2^10 - 1))/(2 - 1)`
= 2(1024 – 1)
= 2046
APPEARS IN
संबंधित प्रश्न
Which term of the following sequence:
`sqrt3, 3, 3sqrt3`, .... is 729?
For what values of x, the numbers `-2/7, x, -7/2` are in G.P?
Find the sum to n terms of the sequence, 8, 88, 888, 8888… .
If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that `a^(q - r) b^(r-p) c^(p-q) = 1`.
The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio `(3 + 2sqrt2) ":" (3 - 2sqrt2)`.
Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P2Rn = Sn
Show that the sequence <an>, defined by an = \[\frac{2}{3^n}\], n ϵ N is a G.P.
Find:
the 10th term of the G.P.
\[- \frac{3}{4}, \frac{1}{2}, - \frac{1}{3}, \frac{2}{9}, . . .\]
Which term of the G.P.: `sqrt3, 3, 3sqrt3`, ... is 729?
Which term of the G.P. :
\[\frac{1}{3}, \frac{1}{9}, \frac{1}{27} . . \text { . is } \frac{1}{19683} ?\]
The product of three numbers in G.P. is 216. If 2, 8, 6 be added to them, the results are in A.P. Find the numbers.
The sum of three numbers in G.P. is 21 and the sum of their squares is 189. Find the numbers.
Find the sum of the following geometric progression:
4, 2, 1, 1/2 ... to 10 terms.
Find the sum of the following geometric series:
x3, x5, x7, ... to n terms
Find the sum of the following series:
9 + 99 + 999 + ... to n terms;
The common ratio of a G.P. is 3 and the last term is 486. If the sum of these terms be 728, find the first term.
The fifth term of a G.P. is 81 whereas its second term is 24. Find the series and sum of its first eight terms.
If S1, S2, ..., Sn are the sums of n terms of n G.P.'s whose first term is 1 in each and common ratios are 1, 2, 3, ..., n respectively, then prove that S1 + S2 + 2S3 + 3S4 + ... (n − 1) Sn = 1n + 2n + 3n + ... + nn.
Prove that: (91/3 . 91/9 . 91/27 ... ∞) = 3.
The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers.
If a, b, c, d are in G.P., prove that:
(a2 + b2), (b2 + c2), (c2 + d2) are in G.P.
If \[\frac{1}{a + b}, \frac{1}{2b}, \frac{1}{b + c}\] are three consecutive terms of an A.P., prove that a, b, c are the three consecutive terms of a G.P.
Find the geometric means of the following pairs of number:
2 and 8
If in an infinite G.P., first term is equal to 10 times the sum of all successive terms, then its common ratio is
If A be one A.M. and p, q be two G.M.'s between two numbers, then 2 A is equal to
Given that x > 0, the sum \[\sum^\infty_{n = 1} \left( \frac{x}{x + 1} \right)^{n - 1}\] equals
The two geometric means between the numbers 1 and 64 are
For the G.P. if a = `2/3`, t6 = 162, find r.
Mosquitoes are growing at a rate of 10% a year. If there were 200 mosquitoes in the beginning. Write down the number of mosquitoes after 10 years.
For a G.P. if a = 2, r = 3, Sn = 242 find n
Determine whether the sum to infinity of the following G.P.s exist, if exists find them:
`1/2, 1/4, 1/8, 1/16,...`
The midpoints of the sides of a square of side 1 are joined to form a new square. This procedure is repeated indefinitely. Find the sum of the areas of all the squares
Select the correct answer from the given alternative.
If common ratio of the G.P is 5, 5th term is 1875, the first term is -
Answer the following:
Find three numbers in G.P. such that their sum is 35 and their product is 1000
Answer the following:
If p, q, r, s are in G.P., show that (p2 + q2 + r2) (q2 + r2 + s2) = (pq + qr + rs)2
