Question

find the maximum sum of the a.p. 30,27,24,21

Answer 1

$sn = \frac{n}{2} (2a + (n - 1) \times d) \\ = \frac{4}{2}(2 \times 30 + (4 - 1) \times - 3 \\ = 2(60 + 3 \times - 3) \\ = 2 \times (60 - 9) \\ = 2 \times - 51 \\ = - 102 \\ the \: sum \: of \: given \: ap \: is \: - 102$

Answer 2

d=-3
a=30
An=a+(n-1)d
therefore 21=30+(n-1)(-3)
n=4
Sn=n/2(a+An). (last term =21)
therefore S4=4/2(30+21)
therefore S4=102

I hope this will help you