K square -4 k-8-2 k square -2 k-6-3 k square -4 k-4

Lets take each term as a, a2 and a3. if they are terms of an AP then a2 a = a3 a2 2k^ + 2k + 6 (k^ + 4k + 8)= 3k^ + 4k + 4 (2k^ + 2k + 6) 2k^ k^ + 2k ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

General Form of an AP

K square +4 k...

Question

Determine $k$ , so that $k^{2} + 4 k + 8, 2 k^{2} + 3 k + 6$ and $3 k^{2} + 4 k + 4$ are three consecutive terms of an AP.

Open in App

Solution

Let's take each term as $a_{1} = k^{2} + 4 k + 8$ , $a_{2} = 2 k^{2} + 3 k + 6$ and $a_{3} = 3 k^{2} + 4 k + 4$ .

if they are terms of an AP then $a_{2} - a_{1} = a_{3} - a_{2}$

$(2 k^{2} + 3 k + 6) - (k^{2} + 4 k + 8) = 3 k^{2} + 4 k + 4 - (2 k^{2} + 2 k + 6)$

$2 k^{2} - k^{2} + 3 k - 4 k + 6 - 8 = 3 k^{2} - 2 k^{2} + 4 k - 2 k + 4 - 6$

$k^{2} - k - 2 = k^{2} + 2 k - 2$

$3 k = 0$
$k = 0$

Suggest Corrections

Similar questions

Determine k so that k^2 + 4k + 8,2k^2 + 3k + 6,3k^2 + 4k + 4 are three consecutive terms of an A.P.

Q. Find

K

K^{2} + 4 K + 8, 2 K^{2} + 3 K + 6

and

3 K^{2} + 4 K + 4

are any

3

consecutive terms of

A . P

Q. If

2 k, 4 k (3)

, and

(4 k + 4)

are in AP. Find k.

The square of a positive integer cannot be a form/(, K is an integer) :

(a) 3 k + 1

(b) 4 k + 2

(d) 5 k + 4

Good evening,

My doubt is

The square of a positive integer CANNOT be of the form (k is an integer):

(a) 3k + 1 (b) 4k + 2 (c) 5k + 1 (d) 5k + 4

Join BYJU'S Learning Program

Determine k, so that k2+4k+8,2k2+3k+6 and 3k2+4k+4 are three consecutive terms of an AP.

Let's take each term as a1=k2+4k+8, a2=2k2+3k+6 and a3=3k2+4k+4. if they are terms of an AP then a2−a1=a3−a2 (2k2+3k+6)−(k2+4k+8)=3k2+4k+4−(2k2+2k+6) 2k2−k2+3k−4k+6−8=3k2−2k2+4k−2k+4−6 k2−k−2=k2+2k−2 3k=0k=0

Determine $k$ , so that $k^{2} + 4 k + 8, 2 k^{2} + 3 k + 6$ and $3 k^{2} + 4 k + 4$ are three consecutive terms of an AP.