Question

The seventh team of an AP is four times it's second term and twelfth term is 2 more than three times of its fourth term. Find the progression

Answer 1

Answer:

a7 = 4(a2) = a+6d

a12 = 3(a4)+2 = a+11d

therefore a+6d = 4(a2) = 4(a+d) = 4a+4d =

                a+6d = 4a+4d

                a+6d-4a-4d = 0

                -3a+2d = 0 -------- (1)

and a+11d = 3(a4)+2 = 3(a+3d)+2 = 3a+9d+2

      a+11d = 3a+9d+2

      a+11d-3a-9d-2 = 0

      -2a+2d-2 = 0

      -2a+2d = 2 --------- (2)

by subtracting (1) and (2),we get

      -3a+2d = 0

   -  (-2a+2d) = 2

    = -3a+2a+2d-2d = -2

    = a = -2

by substituting a in (1),we get

    -3(-2)+2d = 0

    6+2d = 0

    2d = -6

    d = -6/2

    d = -3

therefore the AP = a, a+d, a+2d, a+3d, a+4d..........

                            = -2,  (-2)+(-3),  (-2)+2(-3),  (-2)+3(-3),  (-2)+4(-3).........

                            = -2, -5, -8, -11, -14.............

       

                           

Step-by-step explanation: