Question

If ( aⁿ + bⁿ + cⁿ + dⁿ / aⁿ-1+ b ⁿ-1 + cⁿ-1 + dⁿ-1) is the AM between a and b then find the value of n .

Answer 1

Concept :
a, A1, A2, A3, ...........An , b are in AP then
A1, A2, A3, ......An known as Arithmetic mean between a and b .

(n+2)th term = b { because we started after a and end before b of AM series}
so,
a + (n+2-1)d = b
where d is common difference.

(n+1)d = (b-a)
d = (b-a)/(n+1)
so, An = a + nd = a + n(b-a)/(n+1)
= {an+a +nb-na}/(n+1)
= (nb + a)/(n+1) ____________(1)

now given, (aⁿ + bⁿ)/{a^(n-1)+b^(n-1)} is the AM between a and b . _________(2)

eqns (1) and (2) both are equal when n = 1
An = (nb+a)/(n+1)
for n =1
A1 = (a + b)/2

and An = (aⁿ+bⁿ){a^(n-1)+b^(n-1)}
for n = 1
A1 = (a¹ + b¹)/(a^0+b^0)
= (a + b)/2

so, answer is n = 1

Answer 2

Hi friend ...

Hope it is helpful..