Question

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If 4m+8, 2m2 + 3m +6, 3m2 +4m + 4 are three consecutive terms of an A.P. find
m.
should be taken so that their sum is​

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.