Question

the 8 th term of arithmetic progression is zero . prove that it's 38 th term is triple of its 18 th term

Answer 1

The 8th term of an Ap is 0, so

a+7d =>a=-7d

38th term = a+37d=>-7d+37d=30d

18th term =a+17d=>-7d+17d=10d

Therefore,38th term is trice the 18th term

Hence,proved

Hope it helped...

Answer 2

Hi

Here is ur answer

It is given that the 8th term i.e, a8 = 0

a=a, d=d.

From the equation,  

$an = a + (n-1)d$

    a8 = a + (8-1)d

    0 = a +7d

Bring 7d to left side

Therefore,          a = -7d

From the question,          a38 = 3(a18)

   a38 = a +(38 - 1)d

          = a + 37d

Substitute a = -7d in the above ....

 So,   = -7d + 37d

   Therefore,    a38 = 30d

Then....   a18 = a +(18-1) d

                     = a+17d

Substitute a=-7d in the above.....

  So,   = -7d +17d

  Therefore,  a18 = 10d

Hence,    30d ( from a38) = 3(10d) (from a18)

               A38 = 3(A18)...

Thus verified

Hope u can understand...