About Numbers?

Question

The sum of first fifty positive even integers is 2550. What is the sum of even integers from 102 to 200 inclusive?

Answer 1

Given sum of first fifty even integers = 2550
sum of even number (102 to 200)
= 100*50 + 2550
= 7550

Answer 2

To solve it only using the fact that the first fifty positive even integers is 2550 (from 2 to 100), you can see that going from 102 to 200 is like counting 100 exactly 50 times and adding the difference of all the even integers from 2 to 100:

100 * 50 + 2550 = 5000 + 2550 = 7550.

Answer 3

Ans=100*50+2550=7550
sum of 1st 50 even integers means the sum of even integers from 2 to 100.

we can write 102 as (100+2)
So, the sum of even integers 102 to 200 is...

=(100+2)+(100+4)+........+(100+98)+(100+100)
=100*50+(2+4+..........+98+100)
=5000+2550
=7550

Answer 4

This is just an arithmetic progression with first term as 102 ,last term 200 and common difference = 2.

Summation of such a series is given by the formula:

Sn = (n/2)(2a + (n-1)d)

Where Sn = sum of first n terms of the progression
n = no of terms of the progression
a = first term of the progression
d = common difference.

For our case n = 50
a = 102
d = 2

So Sn = (50/2)(2*102 + (50-1)2)
= 25(204 + 98)
= 7550 which is the answer..

Answer 5

All you need is primary school arithmetic (and some borrowed genius from young Gauss)...

Using pairwise addition we have 2+100=102, 4+98=102, 6+96=102 and so on.

So we have 25 pairs of 102:

25 x 102 = 2550

Similarly for 102 to 200

102+200=302, 104+198=302 and so on

Therefore we have

25 x 302 = 7550

From the wiki:
A famous story, and one that has evolved in the telling, has it that in primary school [Gauss'] teacher, J.G. Büttner tried to occupy pupils by making them add up the integers from 1 to 100. The young Gauss produced the correct answer within seconds by a flash of mathematical insight, to the astonishment of all. Gauss had realized that pairwise addition of terms from opposite ends of the list yielded identical intermediate sums: 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101, and so on, for a total sum of 50 × 101 = 5050

Answer 6

Sum of integers 1-200:
200 * 201 / 2 = 20100

Sum of integers 1-101:
101 * 102 / 2 = 5151

Sum of integers 102-200:
20100 - 5151 = 14949

Since there are 99 numbers 102 - 200, and there is one more even number, that means:
50 even numbers
49 odd numbers

14949 / 99 * 50 = 7550 even numbers

or

(14949 + 1) / 2 = 7550 even numbers

Answer 7

It's a progression question.
Even numbers? That means the difference is 2.
Use the sum formula, S.n=(n/2)[2a + (n-1)d]
200-102=98
98/2=49+1(including 102)=50
S.50=(50/2)[2(102) + (50-1)2]
S.50=25[302]
S.50=7550.

I'm not sure, though. I suck in progression and I do not really understand your question. Plus I didn't really use your S.50=2550.

Answer 8

Given(2+4+........+100)=2550
(102+104+.........+200)=(50*100)+2550
=7550

Answer 9

100*50+2550=5000+2500
=7500

Answer 10

100*50+2550