Answer :

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……..

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