find the first sum of first 50th even natural numbers of the arithmetic progression ?
Answers
Answer:
The sum of the first 50 even natural numbers is 2550.
Step-by-step explanation:
That is Sn=2+4+6+−−+50
This is an arithmetic series and we have formula for this as, Sn=n2[a+(n−1)d]
To apply this formula, we should know the values of the first number ‘a’, common difference ‘d’ and the number of terms ‘n’. We will get these values in the given series itself. Then substitute the values in the formula of Sn. After simplification of this we will get the required sum.
Complete step-by-step answer:
Write the given series, Sn=2+4+6+−−+50
By seeing the series we can write the values of
a=2,n=50,d=a2−a1
=4−2=2
We are using the formula to find the sum of the series is
Sn=n2[a+(n−1)d]
Now substitute the values of a, n and d to the above formula,
Sn=502[2(2)+(50−1)2]
=50×22[2+49]=50(51)=2550.
Hope this will help you.
Answer:
here's explanation in attachment
