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If a1, a2, a3 ....... an are positive real numbers whose product is a fixed number ‘c’, then the minimum value of a1 + a2 ..... + an–1 + 2an is(a) n (2c)1/n (b) (n + 1)c1/n (c) 2nc1/n (d) (n + 1) (2c)1/n |
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Answer» (c) n(2c)\(\frac1n\) For positive real numbers, AM > GM ⇒ \(\frac{(a_1+a_2+a_3+.......+2a_n)}{n}\) > (a1.a2.a3.......2an)\(\frac1n\) ⇒ (a1 + a2 + a3 + ......+ 2an) > n(c.2)\(\frac1n\) = n(2c)\(\frac1n\) Hence the least value is n(2c)\(\frac1n\) |
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