Find the sum of first 50 integers and find the first 50even numbers and find the relation between them
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Answered by
1
Answer:
The difference between 2 consecutive numbers is d=4−2=2 . Here, last term i.e. Tn=T50=100 . So, now we have all the data needed for summation of 50 even integers. Thus, summation of the first 50 even positive integers is 2550.
Answered by
2
Answer:
As we know nth term, a
n
=a+(n−1)d
& Sum of first n terms, S
n
=
2
n
(2a+(n−1)d), where a & d are the first term & common difference of an AP.
The sequence of the first 50 even positive integers is given by 2,4,6,...
The above sequence has a first term equal to 2 and a common difference d=2.
We use the nth term formula to find the 50th term.
a
50
=2+2(50−1)=100
We now the first term and last term and the number of terms in the sequence, we now find the sum of the first 50 terms
S
50
=
2
50×(2+100)
=2550
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