Question

If A.M., G.M. and H.M. of first and last terms of the series 100, 101, 102,…. n –1, n are the terms of the series it self then the value of n is (100 < n £ 500 )

• A. 200
• B. 300
• C. 400
• D. 500

Answer 1

Answer: (C)

400

Answer 2

AM = $\frac { 100 + n } { 2 }$ , G.M. = $10 \sqrt { n }$ , H.M. = $\frac { 200 \mathrm { n } } { 100 + \mathrm { n } }$

for AM to be integer, n must be even

for GM to be integer, n must be perfect square

So n = 4k² HM = $\frac { 200 \mathrm { k } ^ { 2 } } { 25 + \mathrm { k } ^ { 2 } }$

for HM to be integer k² must be divisible by 25 also 100 < n £ 500

k² = 50, 75, 100, 125

Hence n = 400 the only which satisfies