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Answer :
For a natural number m greater than 1,Triplets are in the form 2m, m2 – 1, m2 + 1
(i) Let \(m^2 – 1 = 6\)
\(\Rightarrow m^2 = 6 + 1 = 7\)
So,here we observe that value of m is not integer.
Now, let us try for \(m^2 + 1 = 6\)
\(\Rightarrow m^2 = 6 – 1 = 5\)
So, here too we have no integer value of m.
Now, let 2m = 6
\(\Rightarrow m = 3\; which\; is \;an\; integer.\)
Other number are:
\(m^2 – 1 = 32 – 1 = 8\) and \(m^2 + 1 = 32 + 1 = 10\)
Hence, the required triplets are 6, 8 and 10.
(ii) Let \(m^2 – 1 = 14\)
\(\Rightarrow m^2 = 1 + 14 = 15\)
So, we observe that the value of m will not be an integer.
Now take 2m = 14
\(\Rightarrow\) m = 7 which is an integer.
The other numbers of triplets are 2m =\( 2 \times 7\) = 14
\(m^2 – 1 = (7)2 – 1 = 49 – 1 = 48\)
and \(m^2 + 1 = (7)2 + 1 = 49 + 1 = 50\)
Hence the triplets are 14, 48 and 50.
(iii) Let, 2m = 16
\(\Rightarrow\)m = 8
The other numbers are
\(m^2 – 1 = (8)2 – 1 = 64 – 1 = 63\)
\(m^2 + 1 = (8)2 + 1 = 64 + 1 = 65\)
Hence the triplets are 16, 63 and 65
(iv) Let 2m = 18
\(\Rightarrow\) m = 9
The other triplets are:
\(m^2 – 1 = (9)2 – 1 = 81 – 1 = 80\)
and \(m^2 + 1 = (9)2 + 1 = 81 + 1 = 82\)
Hence we have the triplets as 18, 80 and 82.