Question

Find the maximum sum of the AP 40,38,36,34,32..................

Answer 1

When the A.P,$40,38,36,34,32....$reaches $0$ we observe that on adding it does not contribute anything to the sum and the sum even starts decreasing once it reaches $-2$.Therefore,the sum has to be evaluated till $2$.Now,$a=40\\d=-2\\n=20\\S=\frac{a}{2}[2a+(n-1)d]\\=\frac{20}{2}[2\cdot40+(20-1)(-2)]=10(80-38)=420$

Answer 2

AP series : 40, 38, 36, 34, 32, ...

⇒a1 = 40

⇒ d = 38 - 40 = -2

⇒ For max sum, the last term would be 0

Find the nth term:

an = a1 + (n - 1)d

an = 40 + (n - 1)(-2)

an = 40 -2n + 2

an = 42 - 2n

Find the number of terms:

an = 42 - 2n

0 = 42 - 2n

2n = 42

n = 21

Find the sum:

Sn = n/2 (a1 + an)

Sn = 21/2 (40 + 0)

Sn = 420

Answer: The maximum sum is 420