Solution -: In this problem we aill apply AM -GM to the numbers a1, a2, a3,… ,an₋₁, 2an
(a1+a2+a3+… +an₋₁+2an) / n ≥ (a1*a2*a3*… *an₋₁*2an)^(1/n)
There is equality in AM-GM when all quantities are equal
→ (a1+a2+a3+… +an₋₁+2an) / n ≥ (2c)^(1/n) (equality if a1=a2…=an₋₁=2an)
→ a1+a2+a3+… +an₋₁+2an ≥ n(2c)^(1/n) so we will see that minimum value of
a1+a2+a3+… +an₋₁+2an is n(2c)^(1/n) so we got answer
n(2c)^(1/n) .