Obtain the sum of all positive integers up to 1000, which

D (which is clearly established by now;) )

The way I figured it was this way.

Multipe of 5 is 5, 10, 15, 20, etc. Those not divisible by 2 are 5, 15, 25, 35, etc.

So, there will be 1 number to include for every 10, so 1000 numbers / 10 = 100 numbers to include in the sum.

If you look at 5 + 995 = 1000, 15 + 985 = 1000, 25 + 975=1000, 35 + 965 = 1000...495+505=1000, You do this, then you have 50 pairs that add up to be 1000, so 50 * 1000 = 50,000 (i.e. D)

I tend to see things in patterns rather than formulas.

Hi!

For those who asked to explain the formula:

For arithmetic progression,

Sn = n*(a1+an)/2

Our series is 5, 15, 25, … , 995. (a1=5, an=995, d=10).

Only, we need to calculate n. In order to find it, we can use the formula an = a1+(n-1)*d – just put in numbers and solve for n.

There are 200 no's divisible by 5 upto 1000 and out of this only 100 nos are not divisible by 2. So the sum we need is (5+15+....+995) or 5(1+3+5+....199). The sum of consequtive n integers is n^2. Therefe for sum of 100 odd nos is 100^2.

Ans = 5 (100^2)=50000 .D

In your example, 1,3,5,..,199 are not consecutive integers (and I don't see where this results "n^2" come from).

What you can write is that the number we look for is

\(\sum_{k=1 and k is odd}^{200} 5*k = \sum_{i=0}^{99} 5(2i+1) = 5 \left( \sum_{i=0}^{99} 2i + \sum_{i=0}^{99} 1 \right) = 5 \left( 2 \sum_{i=0}^{99} i + 100 \right) = 5 \left( 2 \frac{99*100}{2} + 100 \right) = 5 \left( 9,900 + 100 \right) = 50,000\)

5,15,25,.....,995
total terms = 100

sum = n/2 [2a + (n-1)d]

n = 100
a = 5
d = 10

Sum = 50000

D is the answer.

5+15+25+...+995=
1*5+3*5+5*5+...+199*5=
5*(1+3+5+...+199)

Sum of first n odd numbers: \(n^2\)

#of odd numbers: \(n=\frac{(199-1)}{2}+1 = \frac{199-1+2}{2}=100\)

5*100*100=50000

My approach:

Step 1: Find out the sum of all integers divisible by 5. Number of such terms= \(\frac{(1000-0)}{5} + 1= 201\). Sum of all such integers= \(201*500\) (using the sum formula of terms in Arithmetic Progression).. (i)
Step 2: Find out the sum of all integers divisible by 10: Number of such terms= \(\frac{(1000-0)}{10}= 101\). Sum of all such integers= \(101*500= 101*500\) .. (ii)

Now, the multiples of 10 are the ones which are divisible by both 5 and 2. Therefore, we need to subtract (ii) by (i) to arrive at the terms divisible only by 5 and not by 2

Step 3: \(201*500 - 101*500\)
= \(500*(201-101)\)
= \(500*100\)
= \(50,000\). Answer.

Thus, the answer is D.

This is the quickest approach IMO. you really dont need anything else!
+1

Sequence is: 5,15,25,....995

Common difference d = 10
Number of elements: ((995-5)/10)+1=100
Average: (first+last)/2 = (5+995)/2 = 500

Sum= Number of elements * Average = 100 * 500 = 50000

Ans: "D"

100 possibilities, evenly spaced by 10. Since its an evenly spaced set, the median (500) is also the mean. Find the median: will be the average between the 50th observation (495) and the 51st observation (505), which is 500. Average * number of observations == total. 500*100 is 50,0000.

Can u explain in brief why do you multiply by 500. Thank you

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For every evenly spaced set (a.k.a. arithmetic progression), the sum equals to (mean)*(# of terms). 500 is the mean there: (mean)=(first+last)/2=(5+995)/2=500.

Hope it's clear.

Total Multiples of 5 from 1 through 1000 = 1000/5 = 200

Half of these 200 multiples of 5 will be even (i.e. divisible by 2) and remaining half will be odd multiples of 5

i.e. Question : 5+15+25+......+955=?

Since it's an Arithmetic Progression (In which difference between any two consecutive terms remain constant) in which sum of first and last term = Sum of second and Second last term = and so on...

100 such numbers i.e.e 50 such pairs and sum of each pair = 5+955 = 1000

i.e. Sum of all pairs = (100/2)*(5+955) = 50*1000 = 50,000

Answer: Option D

The question is basically asking us the sum of the AP series
5+15+......995
Number of terms can be calculated from A(n)= A+(n-1)D
So N=100
Sum =100/2 * [5+995] = 50,000
hence D

Let's first try to figure out what the problem's actually asking us.
Numbers divisible by 5 mean the set : {5,10,15,....,1000}
Numbers divisible by 5 and 2 mean the set : {10,20,30,...,1000}
OUR CRITERION: divisible by 5 and not divisible by 2
So, the set we need to work with is : {5,15,25,35,...995}
This is a generic arithmetic progression problem, where the common difference is 10 and number of numbers is 100*, so we can easily apply the formula,

s=n/2*{2a+(n-1)d} where s=sum of numbers, n=number of numbers a=first term and d=common difference
So, s=100/2*{2*5+(100-1)*10}=50,000 ...OPTION D

*number of numbers is 100 because between 1-10, we have only one number, i.e. 5, now multiply it to our total 1-1000, i.e. 100.

The sum of positive integers which are less than 1000 and divisible by 5 but not by 2.
The numbers divisible by 5 are written in their general form: 5*k.
The numbers are also divisible by 2 if k is an even number.
Hence k only takes odd numbers.
The highest odd number k can take so that the number 5*k is less than 1000 is 199.
Hence the sum of the numbers :
5*( 1+3+5+............199)

= Sum of first n odd numbers starting from 1 is \( n^2\)

5*(100*100)

= 50000

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