Q12 of 84 Page 5

For what value of k will k + 9, 2k – 1 and 2k + 7 are the consecutive terms of an A. P.? (CBSE 2016)

We know that three consecutive terms (Say a1, a2 and a3 ) are in AP. If the differences of successive terms are equal.

i.e. a2 - a1 = a3 - a2

In this case

a1 = k + 9

a2 = 2k - 1

a3 = 2k + 7

For these terms to be in AP

2k - 1 - (k + 9) = 2k + 7 - (2k - 1)

k - 10 = 8

k = 10 + 8

k = 18

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