Problem Solving

The sum of the first 50 positive even integrs is 2550. What is the sum of the even integrs from 102 to 200 inclusive ??


5100
7550
10100
15500

Ans B

Is there a formula I can use to find the sum on concective numbers??

Sum of N consecutive numbers is given by n(n+1)/2.

But you need not to know the formula for solving this question.
Sum of first 50 even integers is given to be 2550. Now you have to identify that all the even integers from 102 to 200 is just the list of first 50 even integers with 100 added to each element.
2 => 102
4 => 104
..
..
100 => 200

So the sum would be 50*100 + 2550 = 7550

For example the sum of the first 100 integers is (first + last)*(one half of the number of terms)
(1+100)(100/2)=101*50=5050

we can see that the sum of the first 50 even poz integers is 2550.
divide 2550 by half of the terms(25)--2550/25=102
102 is the sum of the first and last even term or 2+100

The question asks about 102 and 200
so we have 102+200=302 (the sum)
302*25=7550

I hope this helps

gabriela13 wrote:For example the sum of the first 100 integers is (first + last)*(one half of the number of terms)
(1+100)(100/2)=101*50=5050

Great approach.

What we're really using here is the average formula.

Average = sum of terms / # of terms.

For a set of consecutive numbers, this always works out to:

Avg = (1st term + last term)/2

Since the sum = avg * # of terms, we can derive:

Sum of a set of consecutive integers = (1st term + last term)/2 * # of terms

So, in this question:

Sum = (102 + 200)/2 * 50 = 151 * 50 = 7550

Note that on this particular question we really didn't need to be so precise. Glancing at the choices, we see that they're pretty far apart. If we recognize that 150 is approx the average (it's about halfway between 102 and 200) and that there are about 50 even numbers from 102 to 200, we could have just estimated 50*150 = 7500 and then choice (b) is the only one even close.

It could be also solved as follows:

(sum of even integers from 2-200) - (sum of even integers from 2-100)

Formula to find sum of even integers = n(n+1)

where n is the number of even terms.

n = number of even terms = (first even term + last even term)/2 - 1

so number of even terms from 2-200 is
n = 2 + 200/2 - 1 = 100

Therefore, sum of even integers from 2-200: 100(101) = 10,100

sum of even integers from 2-100: 50(51) = 2,550
where n = (2+100)-1 = 50

sum from 102-200 = 10,100 - 2,550 = 7,550

The specified problem is on Arithmetic Progressions (AP).

The sum of the first n terms of an a.p. is given by :
n/2 (2a + (n − 1)d)

The nth term of an a.p. is given by:
a + (n − 1)d

where the first term is a, the common difference is d and n is the total number of elements.

For the specified problem, substitute a=102, d=2 and n=50 and the answer would come to 7550.

A helpful document on Arithmetic and Geometric Progressions (AP & GP) can be found below :

https://www.mathcentre.ac.uk/resources/l ... ssions.pdf

Hope this helps.

-Shaz

Stuart Kovinsky wrote:

gabriela13 wrote:For example the sum of the first 100 integers is (first + last)*(one half of the number of terms)
(1+100)(100/2)=101*50=5050

Great approach.

What we're really using here is the average formula.

Average = sum of terms / # of terms.

For a set of consecutive numbers, this always works out to:

Avg = (1st term + last term)/2

Since the sum = avg * # of terms, we can derive:

Sum of a set of consecutive integers = (1st term + last term)/2 * # of terms

So, in this question:

Sum = (102 + 200)/2 * 50 = 151 * 50 = 7550

Note that on this particular question we really didn't need to be so precise. Glancing at the choices, we see that they're pretty far apart. If we recognize that 150 is approx the average (it's about halfway between 102 and 200) and that there are about 50 even numbers from 102 to 200, we could have just estimated 50*150 = 7500 and then choice (b) is the only one even close.

great stuff
thanks

moneyman wrote:The sum of the first 50 positive even integrs is 2550. What is the sum of the even integrs from 102 to 200 inclusive ??


5100
7550
10100
15500

Ans B

Is there a formula I can use to find the sum on concective numbers??

The formula given above is useful. But on a more intricate problem, the below approach which uses number properties may also be beneficial

The sereis of numbers we want to add is :
102, 104, 106, .......................196, 198, 200.

Start working your way from both extremeties of this sereis of numbers.
102+ 200 = 302
104 +198 = 302
106 +196 = 302

This is because for every increase of 2 from the left, there is a commensurate decrease from the right of 2. Therefore, Sum remains unchanged.

How many such PAIRS of numbers do we have? 25; since there are 50 even numbers between 102 & 200.

Therefore, Sum of given sereis = (302) (25) = 7550 = B

One can see that there should be a very easy way to solve this

2550 is the sum of the first 50 even numbers (positive)

We need to use 2550 ( that's why it is given in the first place )

We know that there are 50 even integer from 102 to 200 inclusive

if we add from 102 to 200 what will we get ?????????

102+104+......................200

Its nothing but (100+2)+(100+4)+............+(100+100)

and we have 50 hundreds if we take the twos we have 2(1+2+..+50) =2550

Hence the answer is 50*100+2550 = 7550

where did you get "Sum of N consecutive numbers is given by n(n+1)/2." ?

i believe that formula show be (first term + last term) / 2.

simplyjat wrote:Sum of N consecutive numbers is given by n(n+1)/2.

But you need not to know the formula for solving this question.
Sum of first 50 even integers is given to be 2550. Now you have to identify that all the even integers from 102 to 200 is just the list of first 50 even integers with 100 added to each element.
2 => 102
4 => 104
..
..
100 => 200

So the sum would be 50*100 + 2550 = 7550