Question

41. Find the sum of series 103+101+99+.....49.​

Answer 1

Answer:

As we can see, it's an A.P.

where,

A= 103

D= -2

An=A+(n-1)D

49=103+(n-1)(-2)

-54=(n-1)(-2)

(n-1)=27

n=28

Now, Sn=n/2[2A+(n-1)D]

S28 = 28/2[2(103)+(28-1)(-2)]

S28 = 14[206-54]

S28 = 14*152

S28 = 2128

Answer 2

Answer:

an=n/2(2a+(n-1)d)

Step-by-step explanation:

a=103

d=-2

an=49

49=103-2n+2

54=-2n