AP.........finds AP or GP or HP

Algebra Level 4

If a , b , c a, b, c are in AP, then

a + 1 b c , b + 1 c a , c + 1 a b a + \frac{1}{bc}, b + \frac{1}{ca} , c + \frac{1}{ab}

are in

AP HP AGP GP

Arpit Dhimanj by Arpit Dhimanj , 23, India

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2 solutions

a + 1 b c , b + 1 c a , c + 1 a b , a + \dfrac{1}{bc}, b + \dfrac{1}{ca},c + \dfrac{1}{ab},

= a b c + 1 b c , a b c + 1 a c , a b c + 1 b a = \dfrac{abc+1}{bc},\dfrac{abc+1}{ac},\dfrac{abc+1}{ba}

= ( a b c + 1 ) 1 b c , ( a b c + 1 ) 1 a c , ( a b c + 1 ) 1 b a = (abc+1)\dfrac{1}{bc}, (abc+1)\dfrac{1}{ac}, (abc+1)\dfrac{1}{ba}

= ( a b c + 1 ) a a b c , ( a b c + 1 ) b a b c , ( a b c + 1 ) c a b c = (abc+1)\dfrac{a}{abc},(abc+1)\dfrac{b}{abc},(abc+1)\dfrac{c}{abc}

= ( a b c + 1 a b c ) a , ( a b c + 1 a b c ) b , ( a b c + 1 a b c ) c = (\dfrac{abc + 1}{abc})a, (\dfrac{abc + 1}{abc})b, (\dfrac{abc + 1}{abc})c

= k a , k b , k c = ka,kb,kc where k = a b c + 1 a b c k = \dfrac{abc + 1}{abc} .

Now we know that if a , b , c a,b,c are in AP, then k a , k b , k c ka,kb,kc are also in AP.

Therefore, the given series is in AP.

3 Helpful 1 Interesting 0 Brilliant 0 Confused
Raj Rajput
May 1, 2015

what i prefer for all is assume A.P a=1 , b=2, c=3 put values get answer as we have to save time

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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