Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Book Online Demo
Check Your IQ
Try Test

Courses

Dropper NEET CourseDropper JEE CourseClass - 12 NEET CourseClass - 12 JEE CourseClass - 11 NEET CourseClass - 11 JEE CourseClass - 10 Foundation NEET CourseClass - 10 Foundation JEE CourseClass - 10 CBSE CourseClass - 9 Foundation NEET CourseClass - 9 Foundation JEE CourseClass -9 CBSE CourseClass - 8 CBSE CourseClass - 7 CBSE CourseClass - 6 CBSE Course
Offline Centres

Q.

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=1nfx=120 then find the value of n.


see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya

a

5

b

4

c

3

d

2 

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

Geometric sequence otherwise known as Geometric progression, abbreviated as GP is a type sequence where the ratio between any two consecutive numbers is constant. If (xn)=x1,x2,x3,...is an GP, then
x2x1=x3x1….=r….....(1)
Here r is called common ratio and if we take the first term x1=a , then sum of GP of first n terms is given by,
S = a(rn-1)r-1
We are given in the question the summation value ∑x=1nfx=120which we can write by expansion as;
f(1) + f(2)+...+f(n)=120......(1)
We are also given the following functional equation.
f(x+y) = f(x)f(y).........(2)
We are also given f(1)=3. Let us put y=1 in the above equation to have;
f(1+1) = f(1)f(1)
⇒ f(2) = 3⋅3 = 32
We put x=1,y=2  in equation (2) and use previously obtained f(2)= 32 to have
f(2+1) = f(1)f(2)
⇒f(3) = 3⋅32
⇒f(3) = 33
We can go on putting y=3,4,...n+1 to have values of the function as
f(3)= 34,f(4)= 35...,f(n)= 3n
We put these values in equation (1) to have;
3+32+...+3n=120
We see that in the left hand side of the above step, we have have GP with first term say a=3 and common ratio r=323=3332………=3. We use formula for sum of first n terms of a GP and have;
⇒ 3(3n-1)3-1 = 120
⇒ 32(3n-1-1)=120
⇒ 3n-1=120 × 23+1=81
⇒ 3n-1=34
We equate the exponents of both sides of the above equations to have;
⇒n−1= 4
∴ n= 5
 
Watch 3-min video & get full concept clarity
Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Online Course for IIT JEE
Best Online Course for NEET
Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Online Course for IIT JEE
Best Online Course for NEET
Best Online Course for CBSE
whats app icon