Find the value of n so that a^{n + 1} + b^{n + 1}/aⁿ + bⁿ may be the geometric mean between a and b

Solution:

It is known that G.M. of a and b is √ ab

By the given condition

a^{n + 1} + b^{n + 1}/aⁿ + bⁿ = √ab

By squaring both sides, we obtain

(a^{n + 1} + b^{n + 1})²/(aⁿ + bⁿ)² = ab

⇒ a^{2n + 2} + 2a^{n + 1}b^{n + 1} + b^{2n + 2}

= (ab)(a²ⁿ + 2aⁿbⁿ + b²ⁿ)

⇒ a^{2n + 2} + 2a^{n + 1}b^{n + 1} + b^{2n + 2}

= a^{2n + 1}b + 2a^{n + 1}b^{n + 1} + ab^{2n + 1}

⇒ a^{2n + 2} + b^{2n + 2} = a^{2n + 1}b + ab^{2n + 1}

⇒ a^{2n + 2} - ab^{2n + 1} = ab^{2n + 1} - b^{2n + 2}

⇒ a²ⁿ⁺¹ (a - b) = b²ⁿ⁺¹ (a - b)

⇒ (a/b)²ⁿ⁺¹ = 1 = (a/b)⁰

⇒ 2n + 1 = 0

⇒ n = - 1/2

Thus, the value of n = - 1/2

---

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.3 Question 27

Find the value of n so that a^{n + 1} + b^{n + 1}/aⁿ + bⁿ may be the geometric mean between a and b.

Summary:

We had to find the value of n so that the expression a^{n + 1} + b^{n + 1}/aⁿ + bⁿ may be the geometric mean between a and b