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Question

If a, b, c are three natural numbers in AP and a+b+c=21 then the possible number of values of the ordered triplet (a,b,c) is

A
15
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B
14
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C
13
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D
none of these
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Solution

The correct option is C 13
We know that
b=a+c2 since they are in A.P
Hence 2b=a+c
2(a+b+c)=42
2a+2c+2b=42
2a+2c+a+c=42
3(a+c)=42
a+c=14
Therefore b=7
Now a,b,cϵN, we get
a=1,c=13,
a=2,c=12
:
:
:
a=13,c=1
Hence we will have 13 ordered pairs of (a,b,c) as a=(1,2,...13)
c=(13,12,11...1) and b=7

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