Question

sum of even numbers from 20 to 100​

Answer 1

Answer:

We can solve this using the concept of Sum of ’n’ terms of A.P. (arithmetic progression)

2,4,6,…………………………..,100

Here, a=2 (First term)

d=2 (common difference)

l (an)=100 (last term)

an=a+(n-1)d

100=2+(n-1)2

100=2+2n-2

100=2n

n=100÷2

n=50

Sn=n÷2 (a+l)

S50=50÷2 (2+100)

=25×102=2550.

Sum of all even numbers from 2 to 100 is 2550

Step-by-step explanation:

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Answer 2

Answer

$a = 20 \\ d = 2 \\ an = 100$

$an = a + (n - 1)d \\ 100 = 20 + (n - 1)2 \\ \frac{80}{2} = n - 1 \\ 40 + 1 = n \\ 41 = n$

$sn = 41 \div 2(40 + 40 \times 2) \\ 41 \div 2 \times 40 \times 3 \\ 41 \times 60 \\ 2460$

so this is the answer

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