Question

find the sum of series 103+101+99+...49?

Answer 1

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 ANS.

Answer 2

Answer:

The sum of 103+101+99+...49 is 2128.

Step-by-step explanation:

The given series is

103+101+99+...49

It is an AP. Here first term is 103 and common difference is -2.

The nth term of an AP is

$a_n=a+(n-1)d$

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

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

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

$27=(n-1)$

$27+1=n$

$n=28$

The number of terms in the series is 28.

The sum of n terms of AP is

$S_n=\frac{n}{2}[2a+(n-1)d]$

$S_{28}=\frac{28}{2}[2(103)+(28-1)(-2)]$

$S_{28}=14(152)$

$S_{28}=2128$

Therefore the sum of 103+101+99+...49 is 2128.