Question

If 4m +8, 2m² + 3m + 6, 3m² + 4m + 4 are three consecutive terms of an A.P. find m.

Answer 1

Given:  4m +8, 2m² + 3m + 6,  3m² + 4m + 4 will be consecutive terms of an AP

Here, first term = 4m +8
Second term=  2m² + 3m + 6
third term = 3m² + 4m + 4

Second term -  first term = third term -  second term

2m² + 3m + 6  - (4m +8) = 3m² + 4m + 4 -       (2m² + 3m + 6)
2m² + 3m + 6  - 4m - 8 = 3m² + 4m + 4 -       2m² - 3m - 6
2m² + 3m  - 4m + 6 - 8 =  3m² - 2m²+ 4m - 3m + 4 - 6
2m² - m -2 = m² +m -2

2m² - m² -m -m = -2+2
m² -2m = 0
m² - 2m = 0
m ( m - 2)= 0
m = 0 or m - 2=0

m = 0 or m = 2

Hence, the value of m= 0, 2.

HOPE THIS WILL HELP YOU....

Answer 2

 since they are consecutive terms of an A.P.Common Difference is the Same.

4m-6 -m-2 = 3m-2 -4m+6 

3m-8 = 4-m 

4m = 12

m = 3.

Verification:

m+2 , 4m-6 , 3m-2

5 , 6 , 7 

Therefore m is 3 and common difference is 1.