Find the value of the middle term of the following A.P.: -6, -2, 2, …, 58. (CBSE 2011)

OR

Determine the A.P. whose fourth term is 18 and the difference of the ninth term from the fifteenth term is 30. (CBSE 2011)

nth term of an A.P. is given by a_n=a + (n-1)d

Given: a=-6; d=-2-(-6) =4; a_n=58

Substituting the values of a and d in the formula of a_n

a + (n-1)d=58

⇒ (-6) + (n-1)4=58

⇒(n-1)4=58 + 8=64
⇒n-1=64/4=16

⇒n=17

We see that n is odd.

So,Middle term=th term==9^th term

Middle term = a + (n-1)d =( -6) + (9-1)4=(-6) + 32=26

Or

Let us assume that,

First term of A.P.=a

Common difference=d

We know that nth term of an A.P. is given by a_n=a + (n-1)d

4^th term of A.P.= a + 3d =18(given)…….(1)

Also, 15^th term – 9^th term =30

⇒ (a + 14d)-(a + 8d) =30

⇒ 6d =30

⇒ d=5

Putting the value of d in equation 1

a + 3(5) =18

a + 15= 18

a= 3

So the A.P. is 3,8,13,18,23,28,33……..