English

Find the value of n so that an+1+bn+1an+bn may be the geometric mean between a and b. - Mathematics

Advertisements
Advertisements

Question

Find the value of n so that  `(a^(n+1) + b^(n+1))/(a^n + b^n)` may be the geometric mean between a and b.

Sum
Advertisements

Solution

The geometric mean between a and b = `sqrt"ab"`

⇒ `("a"^("n"+ 1) + "b"^("n" + 1))/("a"^"n" + "b"^"n") = sqrt"ab"`

∴ `"a"^("n"+ 1) + "b"^("n" + 1) = sqrt"ab" ("a"^"n" + "b"^"n")` 

= `"a"^("n"+ 1/2) "b"^(1/2) + "a"^(1/2) "b"^("n" + 1/2)`

or `("a"^("n" + 1) - "a"^("n" + 1/2) "b"^(1/2)) - ("a"^(1/2) "b"^("n" + 1/2) - "b"^("n" + 1)) = 0`

or `"a"^("n" + 1/2) ("a"^(1/2) - "b"^(1/2)) - "b"^ ("n" + 1/2)("a" ^(1/2) - "b"^(1/2)) = 0`

or `("a"^(1/2) - "b"^(1/2)) ("a"^("n" + 1/2) - "b"^ ("n" + 1/2)) = 0`

`"a" ^(1/2) - "b"^(1/2) ≠ 0`

∴ `"a"^("n" + 1/2) - "b"^ ("n" + 1/2) = 0`

or `"a"^("n" + 1/2) = "b"^ ("n" + 1/2)`

or `("a"/"b")^("n"+1/2) = 1 = ("a"/"b")^0`

⇒ `"n"+ 1/2 = 0` 

n = `(-1)/2`

shaalaa.com
  Is there an error in this question or solution?
Chapter 9: Sequences and Series - Exercise 9.3 [Page 193]

APPEARS IN

NCERT Mathematics [English] Class 11
Chapter 9 Sequences and Series
Exercise 9.3 | Q 27 | Page 193

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

If a, b, c, d are in G.P, prove that (an + bn), (bn + cn), (cn + dn) are in G.P.


If a, b, c are in A.P,; b, c, d are in G.P and ` 1/c, 1/d,1/e` are in A.P. prove that a, c, e are in G.P.

 

Show that one of the following progression is a G.P. Also, find the common ratio in case:

4, −2, 1, −1/2, ...


Find:
the ninth term of the G.P. 1, 4, 16, 64, ...


Which term of the G.P. :

\[\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}, \frac{1}{4\sqrt{2}}, . . . \text { is }\frac{1}{512\sqrt{2}}?\]


Find the 4th term from the end of the G.P.

\[\frac{1}{2}, \frac{1}{6}, \frac{1}{18}, \frac{1}{54}, . . . , \frac{1}{4374}\]


The fourth term of a G.P. is 27 and the 7th term is 729, find the G.P.


If a, b, c, d and p are different real numbers such that:
(a2 + b2 + c2) p2 − 2 (ab + bc + cd) p + (b2 + c2 + d2) ≤ 0, then show that a, b, c and d are in G.P.


Find the sum of the following geometric progression:

1, 3, 9, 27, ... to 8 terms;


Find the sum of the following geometric progression:

1, −1/2, 1/4, −1/8, ... to 9 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:

`sqrt7, sqrt21, 3sqrt7,...` to n terms


If S1, S2, S3 be respectively the sums of n, 2n, 3n terms of a G.P., then prove that \[S_1^2 + S_2^2\] = S1 (S2 + S3).


If Sp denotes the sum of the series 1 + rp + r2p + ... to ∞ and sp the sum of the series 1 − rp + r2p − ... to ∞, prove that Sp + sp = 2 . S2p.


If a, b, c are in G.P., prove that \[\frac{1}{\log_a m}, \frac{1}{\log_b m}, \frac{1}{\log_c m}\] are in A.P.


The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an A.P. Find the numbers.


If a, b, c are in G.P., prove that:

a (b2 + c2) = c (a2 + b2)


If a, b, c are in G.P., prove that:

\[a^2 b^2 c^2 \left( \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} \right) = a^3 + b^3 + c^3\]


If a, b, c are in G.P., prove that:

\[\frac{1}{a^2 - b^2} + \frac{1}{b^2} = \frac{1}{b^2 - c^2}\]


If a, b, c, d are in G.P., prove that:

(a2 + b2), (b2 + c2), (c2 + d2) are in G.P.


Insert 6 geometric means between 27 and  \[\frac{1}{81}\] .


Insert 5 geometric means between 16 and \[\frac{1}{4}\] .


Find the geometric means of the following pairs of number:

−8 and −2


If (p + q)th and (p − q)th terms of a G.P. are m and n respectively, then write is pth term.


If A1, A2 be two AM's and G1G2 be two GM's between and b, then find the value of \[\frac{A_1 + A_2}{G_1 G_2}\]


If for a sequence, tn = `(5^("n"-3))/(2^("n"-3))`, show that the sequence is a G.P. Find its first term and the common ratio


Find four numbers in G.P. such that sum of the middle two numbers is `10/3` and their product is 1


Find five numbers in G.P. such that their product is 1024 and fifth term is square of the third term.


A ball is dropped from a height of 80 ft. The ball is such that it rebounds `(3/4)^"th"` of the height it has fallen. How high does the ball rebound on 6th bounce? How high does the ball rebound on nth bounce?


For the following G.P.s, find Sn

0.7, 0.07, 0.007, .....


For a sequence, if Sn = 2(3n –1), find the nth term, hence show that the sequence is a G.P.


If S, P, R are the sum, product, and sum of the reciprocals of n terms of a G.P. respectively, then verify that `["S"/"R"]^"n"` = P


Find: `sum_("r" = 1)^10 5 xx 3^"r"`


Find : `sum_("r" = 1)^oo 4(0.5)^"r"`


Select the correct answer from the given alternative.

The common ratio for the G.P. 0.12, 0.24, 0.48, is –


Answer the following:

For a sequence Sn = 4(7n – 1) verify that the sequence is a G.P.


Answer the following:

Find k so that k – 1, k, k + 2 are consecutive terms of a G.P.


Answer the following:

Find the sum of infinite terms of `1 + 4/5 + 7/25 + 10/125 + 13/6225 + ...`


Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×