Question

Find the sum of first 50 integers and find the first 50even numbers and find the relation between them​

Answer 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.

Answer 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