Question

Find the value of m so that m+2, 4m-6 3m-2 are three consecutive terms of an AP.​

Answer 1

Answer:

The given sequence is an Arithmetic Progression. The difference between the two terms is constant.

t1 = m+2, (first term)

t2 = 4m -6 (second term)

t3 = 3m-2 (Third term)

t2-t1= t3-t2

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

4m-6-m-2 = 3m-2–4m+6

4m-m-6–2= 3m-4m +6–2

3m - 8 = -m +4

3m+m = 8+4

4m = 12

m= 12/4 = 3

m = 3