Question

Find the sum of two middle terms of hte A.P.: $-\frac{4}{3},-1,\frac{-2}{3},-\frac{1}{3},...,4\frac{1}{3}$

Answer 1

Here in the given A.P the first term a , = -4/3
an = 11/3 and d = 1/3

$a_n$ = a + ( n - 1 ) d
$\frac{11}{3}=\frac{-4}{3}+(n-1)\frac{1}{3}$

15 = $\frac{n}{3}-\frac{1}{3}$

$\frac{16}{3}=\frac{n}{3}$

16 = n

Now the middle term =$\frac{n}{2}$ , $\frac{n}{2}$+ 1
= 8th term and 9th term.

=a + 7d + a + 8d

=2a + 15d

= 2(-4/3) + 15(1/3)

=-8/3 + 5

= -8 + 5

= 3

= 7/3