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,
Last term (nth )=431=13/3
First term a=−4/3
Then, difference d=−1−(−4/3)=−1+4/3=(−3+4)/3=1/3
=−2/3−(−1)=−2/3+1=(−2+3)/3=1/3
Therefore, common difference d=1/3
We know that,
Tn=a+(n−1)d
So, 13/3=−4/3+(n−1)(1/3)13/3+4/3=1/3n−1/313/3+4/3+1/3=1/3n(13+4+1)/3=1/3n18/3=1/3n6=1/3nn=6×3n=18
So, middle term is 18/2 and (18/2)+1=9th
and 10th term
Then ,a9+a10=a+8d+a+9d=2a+17d
Now substitute the value of ‘a’ and ‘d’.
=2(−4/3)+17(1/3)
=−8/3+17/3
=(−8+17)/3
=9/3
=3
Therefore, the sum of the two middlemost terms of the A.P is 3