Question

11. Determine k so that
$k+ 4k +8, 2K + 3k + 6, 3k + 4k + 4$
are three consecutive
terms of an AP.​

Answer 1

Answer:

As we know that in an AP the common difference is equal ... Determine k so that k2+ 4k + 8, 2k2 + 3k + 6, 3k2 + 4k + 4 are ... Given, Second term, T2 = 2k^2 + 3k + 6

Answer 2

Answer:

k = 0 .

the following is ur solution......