# If f is a fraction satisfying f (x + y) = f (x)× f (y) for all x, y ∈ N, such that f (1) = 3 and ∑^{n}_{x = 1} f ( x) = 120, find the value of n

**Solution:**

f (x + y) = f (x) × f (y) for all x, y ∈ N and f (1) = 3

Taking x = y = 1 in (1) , we obtain

f (x + y) = f (x) x f (y)

f (1 + 1) = f (1) f (1)

f (2) = 3 x 3

= 9

Similarly,

f (x + y) = f (x) × f (y)

f (1 + 2) = f (1) f (2)

f (3) = 3 x 9

= 27

Also,

f (x + y) = f (x) × f (y)

f (1 + 3) = f (1) f (3)

f (4) = 3 x 27

= 81

Therefore, f (1), f (2), f (3) i.e., 3, 9, 27 forms a G.P. with both the first term and common ratio equal to 3

S_{n} = ∑^{n}_{k = 1}(a)_{k}

It is known that S_{n} = a (1 - r^{n})/(1 - r)

It is given that ∑^{n}_{x = 1} f ( x) = 120

Therefore,

⇒ 120 = 3(3^{n} - 1)/(3 - 1)

⇒ 120 = 3/2 (3^{n} - 1)

⇒ 3^{n} - 1 = 80

⇒ 3^{n} = 81

⇒ 3^{n} = 3^{4}

⇒ n = 4

Thus, the value of n = 4

NCERT Solutions Class 11 Maths Chapter 9 Exercise ME Question 7

## If f is a fraction satisfying f (x + y) = f (x) × f (y) for all x, y ∈ N, such that f (1) = 3 and ∑^{n}_{x = 1} f ( x) = 120, find the value of n

**Summary:**

We found the value of n given that f is a fration satisfying f (x + y) = f (x) × f (y) for all x, y ∈ N, such that f (1) = 3 and ∑^{n}_{x = 1} f ( x) = 120, n = 4