One of the most important question of RS Aggarwal of Arithmetic Progression chapter as it is already asked in previous year paper of 2013 in which we are to find the sum of the middle most terms of the AP -4/3, -1, -2/3,…..,4 1/3.

Class 10, Arithmetic Progression, chapter 5A, question no 19

From the question,

Â LastÂ termÂ (nthÂ )=431â€‹=13/3

First termÂ a=âˆ’4/3

Then, differenceÂ d=âˆ’1âˆ’(âˆ’4/3)=âˆ’1+4/3=(âˆ’3+4)/3=1/3

=âˆ’2/3âˆ’(âˆ’1)=âˆ’2/3+1=(âˆ’2+3)/3=1/3

Therefore, common differenceÂ d=1/3

We know that,

Tnâ€‹=a+(nâˆ’1)d

Â So,Â 13/3=âˆ’4/3+(nâˆ’1)(1/3)13/3+4/3=1/3nâˆ’1/313/3+4/3+1/3=1/3n(13+4+1)/3=1/3n18/3=1/3n6=1/3nn=6Ã—3n=18â€‹

So, middle term isÂ 18/2Â andÂ (18/2)+1=9thÂ

andÂ 10thÂ Â term

Â ThenÂ ,a9â€‹+a10â€‹â€‹=a+8d+a+9d=2a+17dâ€‹

Now substitute the value of ‘a’ and ‘d’.

=2(âˆ’4/3)+17(1/3)

=âˆ’8/3+17/3

=(âˆ’8+17)/3

=9/3

=3â€‹

Therefore, the sum of the two middlemost terms of the A.P is 3