Question

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

Answer 1

Answer:

Step-by-step explanation:

-11,-7 , -3 ......,49 is in AP

first term = a = -11

common difference =d = a2- a1= -7-(-11) = -7 +11 = 4

last term = l = 49

a +(n-1)d =49

-11+(n-1)4 =49

(n-1)4 =49+11

(n-1)4 = 60

n-1 = 60/4

n-1 = 15

n= 15 +1 =16

number of terms are in AP is 16

middle terms or 8th and 9th terms

a8 = a+7d = -11+7*4 = -11+28 = 17

a9 = a +8d = -11+8*4 =-11+32 = 21

Answer 2

In the given A.P,

a (first term) = -11

d (common difference = 4

an (last term) = 49

As we know,

an = a + (n-1)d

49 = -11+(n-1)4

49+11 = (n-1)4

60 = (n-1)4

n-1 = 15

n = 16

So the no. of terms = 16

Since it's even, 8 is the no. of middle most term

a8 = a+7d

= -11+7(4)

= -11+28

= 17

17 is the middle most term