Question

determine K so that 3 k - 2, 2 k square - 5 k + 8 and 4 k + 3 are the three consecutive terms of an ap

Answer 1

Answer:

See if they are consecutive terms of an AP then their common difference will be same.

Answer 2

If the given 3 terms are in A.P., then the common Difference between these terms will be same.

$\tt{a_1}$ = 3k - 2

$\tt{a_2}$ = 2k² - 5k + 8

$\tt{a_3}$ = 4k + 3

Now,

$\tt{a_2 - a_1}$ = $\tt{a_3 - a_2}$

( 2k² - 5k + 8 ) - ( 3k - 2 ) = ( 4k + 3 ) - ( 2k² - 5k + 8 )

2k² - 5k + 8 - 3k + 2 = 4k + 3 - 2k² + 5k - 8

2k² + 2k² - 5k - 3k - 4k - 5k + 8 + 2 - 3 + 8 = 0

4k² - 17k + 15 = 0

By Middle Term Factorisation

4k² - 12k - 5k + 15 = 0

4k ( k - 3 ) - 5 ( k - 3 ) = 0

( 4k - 5 ) ( k - 3 ) = 0

Using Zero Product Rule

4k - 5 = 0 and k - 3 = 0

k = 5 / 4 and k = 3

Hence, the value of k is 5 / 4, 3.