Question

4, k + 4, 8 are in AP find k ?
plz​

Answer 1

${\huge{\bold{\underline{Question:-}}}}$

Question must be ⤵️⤵️

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

${\huge{\bold{\underline{Answer:-}}}}$

if a,b,c are three consecutive terms of an A.P then 2b=a+c

So, 2(2k2+3k+6)=k2+4k+8+3k2+4k+4

4k2+6k+12=4k2+8k+12

6k=8k

∴k=0

Answer 2

Answer:

2

Explanation:

As 4, k + 4, 8 are in A.P.,

Consider 8 be the 3rd term.

a = 4,

d = ?

n = 3

Tn = 8

Tn = a + (n - 1)d

=> 8 = 4 + (3-1)d

=> 8 - 4 = 2d

=> 4 = 2d

d = 2

We can write,

8 - (k + 4) = 2

Multiplying k + 4 by -1,

=> 8 - k - 4 = 2

=> - k + 8 - 4 -2 = 0

=> - k + 2 = 0

=> - k = -2

Cutting the signes,

k = 2