Summation of AP $(= \;\frac{n}{2}\left( {{a_0} + {a_n}} \right))$

Where, n is number of terms, a₀ is first term and a_n is the last term.

It is a decreasing AP.

Thus, maximum sum will be obtained till the last term is positive.

Given AP, 21, $(20\frac{2}{5},\;19\frac{4}{5}, \ldots .)$

C.d. = 20.4 – 21 = -0.6

a_n = a₀ + (n – 1)d = 21 – 0.6(n – 1)

For it to be positive :

21 – 0.6(n – 1) > 0

⇒ 21 > 0.6(n – 1)

⇒ n < 36

∴ n = 35

a_n = 21 – 0.6 × 34 = 0.6