Question

The middle most term (s) of the AP:–11, –7, –3, ..., 49 is: ​

Answer 1

Step-by-step explanation:

Given:

AP:–11, –7, –3, ..., 49

here:

first term, a=-11

last term, l=49.

common difference, d=secend term -first term

=>-7-(-11)=-7+11

d=4.

Formula:

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

An=l

=> 49=-11+(n-1)×4

=> 49+11=(n-1)×4

=> 60/4=n-1

=> 15+1=n

=> 16=n.

There are 16 terms in AP.

So, the middle terms should be 8th or 9th.

(I). 8th term,

=> A8=-11+(8-1)×4

=> -11+7×4

=> -11+28.

=> A8=17.

(ii).9th term,

=> a9=-11+(9-1)×4

=> -11+8×4

=> -11+32

=> A9=21.