Question

9 The maximum sum of the series 100 +98 +96 +... is a) 2500 b) 2550 c) 2050 d) 2555​

Answer 1

Answer:

For Maximum sum

The series must end at positive number

100+98+96+.....+0

Sum = n/2(a+l) where l is the last term

an = a+(n-1)d

0 = 100+(n-1)(-2)

-2n+2+100 = 0

2n = 102

n = 51

sum = 51(100)/2 = 2550

Option B

Answer 2

Answer:

The correct answer is option (b)

Step-by-step explanation:

Given series

  $=100+98+96+...\\=2(50+49+48+...+3+2+1+0+(-1)+(-2)....)$

Clearly the numbers are decreasing so for sum to be maximum we shall add all the the positive terns only

So maximum sum is

      $=2(50+49+48+....+3+2+1)\\=2\times \frac{50}{2}(50+1)\\=50\times 51=2550$

      Formula used :

       $\boxed{\text{S}_n=\dfrac{n}{2}(a+l)}$