Question

determined k so that k²+4k+8, 2k²+3k+6, 3k²+4k+4 are three consecutive terms of A.P.​

Answer 1

GIVEN :-

• Three consecutive terms of AP are [ k² + 4k + 8 ] , [ 2k² + 3k + 6 ] and [ 3k² + 4k + 4 ]

TO FIND :-

• Value of 'k'.

SOLUTION :-

In an AP , Common Difference (d) is same throughout.

Common difference (d) = a(2) - a(1) = a(3) - a(2)

We have ,

• a(1) = k² + 4k + 8
• a(2) = 2k² + 3k + 6
• a(3) = 3k² + 4k + 4

Putting values , we get...

→ (2k² + 3k + 6) - (k² + 4k + 8)

ㅤㅤㅤㅤㅤ= (3k² + 4k + 4) - (2k² + 3k + 6)

→ 2k² + 3k + 6 - k² - 4k - 8

ㅤㅤㅤㅤㅤ= 3k² + 4k + 4 - 2k² - 3k - 6

→ 2k² - k² + 3k - 4k + 6 - 8

ㅤㅤㅤㅤㅤ= 3k² - 2k² + 4k - 3k + 4 - 6

→ k² - k - 2 = k² + k - 2

→ k² - k² - k - k - 2 + 2 = 0

→ -2k = 0

→ k = 0

Hence , value of k is 0.

★ LETS CHECK !

a(1) = k² + 4k + 8 = 8

a(2) = 2k² + 3k + 6 = 6

a(3) = 3k² + 4k + 4 = 4

As Common Difference is same throughout ,

a(3) - a(2) = a(2) - a(1)

Putting values ,

4 - 6 = 6 - 8

-2 = -2 (verified)