Question

Find the sum of the 11th terms of an AP .whose middle term is 30.

Answer 1

 n = 11 
(a1+a11)/2 = 30
(a1+a11)= 60

Sum of AP = n(a1+a11)/2
                  = 11*60/2
                  = 11*30
                  = 330

SUM OF THE 11TH TERMS IS 330

Answer 2

Answer:

330

Step-by-step explanation:

We know for odd term :

Middle term = [ ( n + 1 ) / 2 ]th term :

Given n = 11 :

Middle term = ( 11 + 1 ) / 2 = 6th

We have value of t₆ = 30 :

t_n = a + ( n - 1 ) d

= > t₆ = a + ( 6 - 1 ) d

= > 30 = a + 5 d

= > a = 30 - 5 d ..... ( i )

We know for sum of A.P. :

S_n = n / 2 ( 2 a + ( n - 1 ) d ) )

S₁₁ = 11 / 2 ( 2 a ( 11 - 1 ) d )

Using ( i ) we get :

S₁₁ = 11 / 2 ( 60 - 10 d + 10 d )

S₁₁ = 11 × 30

S₁₁ = 330

Therefore , sum of 11th terms is 330.