Advertisements
Advertisements
प्रश्न
Answer the following:
Find three numbers in G.P. such that their sum is 35 and their product is 1000
Advertisements
उत्तर
Let the three numbers in G.P. be `"a"/"r", "a", "ar"`.
According to the given conditions,
`"a"/"r" + "a" + "ar"` = 35
∴ `"a"(1/"r" + 1 + "r")` = 35 ....(i)
Also, `("a"/"r")("a")("ar")` = 1000
∴ a3 = 1000
∴ a = 10
Substituting the value of a in (i), we get
`10(1/"r" + 1 + "r")` = 35
∴ `1/"r" + "r" + 1 = 35/10`
∴ `1/"r" + "r" = 35/10 - 1`
∴ `1/"r" + "r" = 25/10`
∴ `1/"r" + "r" = 5/2`
∴ 2r2 – 5r + 2 = 0
∴ (2r – 1) (r – 2) = 0
∴ r = `1/2` or r = 2
When r = `1/2`, a = 10
`"a"/"r" = 10/((1/2))` = 20, a = 10 and ar = `10(1/2)` = 5
When r = 2, a = 10
`"a"/"r" = 10/2` = 5, a = 10 and ar = 10 (2) = 20
Hence, the three numbers in G.P. are 20, 10, 5 or 5, 10, 20.
APPEARS IN
संबंधित प्रश्न
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`.
Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is `1/r^n`.
If a, b, c and d are in G.P. show that (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2 .
Find:
the 10th term of the G.P.
\[- \frac{3}{4}, \frac{1}{2}, - \frac{1}{3}, \frac{2}{9}, . . .\]
Which term of the progression 18, −12, 8, ... is \[\frac{512}{729}\] ?
The 4th term of a G.P. is square of its second term, and the first term is − 3. Find its 7th term.
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.
Find the sum of the following geometric series:
0.15 + 0.015 + 0.0015 + ... to 8 terms;
Find the sum of the following geometric series:
1, −a, a2, −a3, ....to n terms (a ≠ 1)
Find the sum of the following geometric series:
x3, x5, x7, ... to n terms
How many terms of the G.P. 3, 3/2, 3/4, ... be taken together to make \[\frac{3069}{512}\] ?
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.
Let an be the nth term of the G.P. of positive numbers.
Let \[\sum^{100}_{n = 1} a_{2n} = \alpha \text { and } \sum^{100}_{n = 1} a_{2n - 1} = \beta,\] such that α ≠ β. Prove that the common ratio of the G.P. is α/β.
Find the sum of the following serie to infinity:
\[1 - \frac{1}{3} + \frac{1}{3^2} - \frac{1}{3^3} + \frac{1}{3^4} + . . . \infty\]
Find the rational numbers having the following decimal expansion:
\[0 . \overline3\]
Three numbers are in A.P. and their sum is 15. If 1, 3, 9 be added to them respectively, they form a G.P. Find the numbers.
If a, b, c, d are in G.P., prove that:
\[\frac{ab - cd}{b^2 - c^2} = \frac{a + c}{b}\]
If a, b, c, d are in G.P., prove that:
(a + b + c + d)2 = (a + b)2 + 2 (b + c)2 + (c + d)2
If a, b, c are in G.P., then prove that:
If a, b, c are in A.P. and a, b, d are in G.P., show that a, (a − b), (d − c) are in G.P.
If S be the sum, P the product and R be the sum of the reciprocals of n terms of a GP, then P2 is equal to
The nth term of a G.P. is 128 and the sum of its n terms is 255. If its common ratio is 2, then its first term is ______.
If x is positive, the sum to infinity of the series \[\frac{1}{1 + x} - \frac{1 - x}{(1 + x )^2} + \frac{(1 - x )^2}{(1 + x )^3} - \frac{(1 - x )^3}{(1 + x )^4} + . . . . . . is\]
The product (32), (32)1/6 (32)1/36 ... to ∞ is equal to
The two geometric means between the numbers 1 and 64 are
The numbers x − 6, 2x and x2 are in G.P. Find x
For the following G.P.s, find Sn
3, 6, 12, 24, ...
Express the following recurring decimal as a rational number:
`2.bar(4)`
The sum of an infinite G.P. is 5 and the sum of the squares of these terms is 15 find the G.P.
Select the correct answer from the given alternative.
Sum to infinity of a G.P. 5, `-5/2, 5/4, -5/8, 5/16,...` is –
Answer the following:
Find the sum of the first 5 terms of the G.P. whose first term is 1 and common ratio is `2/3`
Answer the following:
Which 2 terms are inserted between 5 and 40 so that the resulting sequence is G.P.
Answer the following:
If p, q, r, s are in G.P., show that (p2 + q2 + r2) (q2 + r2 + s2) = (pq + qr + rs)2
Let S be the sum, P be the product and R be the sum of the reciprocals of 3 terms of a G.P. Then P2 R3 : S3 is equal to ______.
In a G.P. of even number of terms, the sum of all terms is 5 times the sum of the odd terms. The common ratio of the G.P. is ______.
If the sum of an infinite GP a, ar, ar2, ar3, ...... . is 15 and the sum of the squares of its each term is 150, then the sum of ar2, ar4, ar6, .... is ______.
