I was looking for solutions online and I found this formula:
d = (thirdLastTerm-3term)/(n-5)
find a & n were easy to understand but in d, I am not getting why (n-5)
It is an AP.
First term =a.
2nd term= a+d
Third term = a+2d.
Similarly, 3rd last term = a+(n-3)d
Now, we are given sum.
sum = n/2 *(2a+(n-1)d)
3 eq, 3 variables.
Now note a few observations before we jump to the problem-
3rd term + 3rd last term = a+2d +a +(n-3)d = 2a+(n-1)d = last term + first term.
Also, sum = n/2 (2a +(n-1)d) = n/2 * (a +[a+(n-1)d]) = n/2 (first term+ last term)
==> n= Sum *2/(first term+last term)
We got n.
a+2d = k1 a+(n-3)d=k2
With n known, we can easily solve this as-
a = k1-2d
==>Subst. in eq.2, we get-
With this we can find a by a=k1-2d
With all three known, the series can be very easily found.
Now as per your query-
Third term= a-2d; Third alst term = a+(n-3)d
What is their difference?
Third Last term - Third term = a+(n-3)d -(a+2d) = (n-5)d
That explains the n-5 there.
Hope this answer helped!! (Feel free to get back in case of any doubts)
Thanks man! very well explained!!
Happy to help dear