for what value of k , the three consecutive terms 2k-7 , k+5, 3k+2 form an A.P
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as d = a2 - a1
=k+5 -(2k+7)
=k+5 - 2k - 7
=12 - k
now,
a3 - a2
= 3k + 2 - ( k+5)
=3k + 2 - k - 5
=2k - 3.....(2)
now, compare 1 and 2
12-k = 2k-3
therefore,
k=5
so our ap is 3 , 10 , 17
=k+5 -(2k+7)
=k+5 - 2k - 7
=12 - k
now,
a3 - a2
= 3k + 2 - ( k+5)
=3k + 2 - k - 5
=2k - 3.....(2)
now, compare 1 and 2
12-k = 2k-3
therefore,
k=5
so our ap is 3 , 10 , 17
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