The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is given by
, then k=
Let the common difference and number of terms of AP be d and n respectively.
Last term of AP = an = l (given)
l = a + (n–1) d
n =
+ 1……………… (1)
S=
(2a + (n–1) d)
S =
+ 1) (2a + (
+ 1 –1) d) …………… using (1)
S=
+ 1) (2a + l –a)
S=
+ 1) (a + l)
(
+ 1) = ![]()
=
–1
= ![]()
D = ![]()
Comparing this with 
We get k =2S
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