Right answer is (a) True
To explain I would say: Consider an AP having n terms in which
First term = a, common difference = d and last term = l.
Then, l = a + (n – 1)d
We may write the given AP as a, a + d, a + 2d …., (l – 2d), (l – d), l
Let Sn be the sum of the first n terms of the AP. Then,
Sn = a + (a + d) + (a + 2d) … + (l – 2d) + (l – d) + l
Writing the above series in reverse order, we get
Sn = (l – 2d) + (l – d) + l … + a + (a + d) + (a + 2d)
Adding the two equations we get,
2Sn = a + l + (a + l) + (a + l) … n times = n(a + l)
Sn = \(\frac {n}{2}\)(a + l)