मराठी

If f is a function satisfying f (x +y) = f(x) f(y) for all x, y ∈ N such that f(1) = 3 and ∑x=1n f(x) = 120, find the value of n.

Advertisements
Advertisements

प्रश्न

If f is a function satisfying f (x +y) = f(x) f(y) for all x, y ∈ N such that f(1) = 3 and `sum_(x = 1)^n` f(x) = 120, find the value of n.

बेरीज
Advertisements

उत्तर

It is given that,

f (x + y) = f (x) × f (y) for all x, y ∈ N … (1)

f (1) = 3

Taking x = y = 1 in (1), we obtain

f (1 + 1) = f (2) = f (1) f (1) = 3 × 3 = 9

Similarly,

f (1 + 1 + 1) = f (3) = f (1 + 2) = f (1) f (2) = 3 × 9 = 27

f (4) = f (1 + 3) = f (1) f (3) = 3 × 27 = 81

∴ f (1), f (2), f (3), …, that is 3, 9, 27, …, forms a G.P. with both the first term and common ratio equal to 3.

It is known that, Sn = `(a(r^n - 1))/(r - 1)`

It is given that, `sum_(x = 1)^nf (x) = 120`

∴ `120 = (3(3^n - 1))/(3 - 1)`

= `120 = 3/2 (3^n - 1)`

= 3n - 1 = 80

= 3n - 1 = 81 = 34

∴ Thus, the value of n is 4.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Sequences and Series - Miscellaneous Exercise [पृष्ठ १४७]

APPEARS IN

एनसीईआरटी Mathematics [English] Class 11
पाठ 8 Sequences and Series
Miscellaneous Exercise | Q 1. | पृष्ठ १४७

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.


The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.


If a and b are the roots of are roots of x2 – 3x + p = 0 , and c, d are roots of x2 – 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q – p) = 17 : 15.


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

4, −2, 1, −1/2, ...


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

\[a, \frac{3 a^2}{4}, \frac{9 a^3}{16}, . . .\]


Show that the sequence <an>, defined by an = \[\frac{2}{3^n}\], n ϵ N is a G.P.


Find : 

nth term of the G.P.

\[\sqrt{3}, \frac{1}{\sqrt{3}}, \frac{1}{3\sqrt{3}}, . . .\]


Find :

the 10th term of the G.P.

\[\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}, . . .\]


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


In a GP the 3rd term is 24 and the 6th term is 192. Find the 10th term.


Find the sum of the following geometric progression:

2, 6, 18, ... to 7 terms;


Find the sum of the following geometric progression:

1, −1/2, 1/4, −1/8, ... to 9 terms;


Evaluate the following:

\[\sum^{11}_{n = 1} (2 + 3^n )\]


Find the sum of the following serie:

5 + 55 + 555 + ... to n 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.


Find the sum of the following serie to infinity:

8 +  \[4\sqrt{2}\] + 4 + ... ∞


Find the rational numbers having the following decimal expansion: 

\[3 . 5\overline 2\]


Find an infinite G.P. whose first term is 1 and each term is the sum of all the terms which follow it.


If a, b, c are in G.P., prove that:

(a + 2b + 2c) (a − 2b + 2c) = a2 + 4c2.


If pth, qth, rth and sth terms of an A.P. be in G.P., then prove that p − q, q − r, r − s 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.


If pq be two A.M.'s and G be one G.M. between two numbers, then G2


If x is positive, the sum to infinity of the series \[\frac{1}{1 + x} - \frac{1 - x}{(1 + x )^2} + \frac{(1 - x )^2}{(1 + x )^3} - \frac{(1 - x )^3}{(1 + x )^4} + . . . . . . is\]


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


For the following G.P.s, find Sn

3, 6, 12, 24, ...


For a G.P. If t4 = 16, t9 = 512, find S10


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


If one invests Rs. 10,000 in a bank at a rate of interest 8% per annum, how long does it take to double the money by compound interest? [(1.08)5 = 1.47]


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


Answer the following:

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


Answer the following:

If for a G.P. first term is (27)2 and seventh term is (8)2, find S8 


Answer the following:

If pth, qth and rth terms of a G.P. are x, y, z respectively. Find the value of xq–r .yr–p .zp–q


Answer the following:

If p, q, r, s are in G.P., show that (p2 + q2 + r2) (q2 + r2 + s2) = (pq + qr + rs)2   


The sum of infinite number of terms of a decreasing G.P. is 4 and the sum of the terms to m squares of its terms to infinity is `16/3`, then the G.P. is ______.


If in a geometric progression {an}, a1 = 3, an = 96 and Sn = 189, then the value of n is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×