The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv

for the 2nd part of the problem, omitting the even integers 25-63, why does the gmat explanation assign d=2?
I understand assigning a=26 and n=19 but not following why d=2.
Any assistance would be appreciated.

thanks

The first part of the problem is straight forward:
Sum of numbers from 1 to 100 = 100*101/2 = 50*101 = 5050

In the second part, you need this sum

26 + 28 + 30 + ... + 62

This is an arithmetic progression with common difference of 2 (between consecutive terms, there is a difference of 2).
26 + (26+2) + (26+2*2) + .... (26+2*18)
a, a+d, a+2d,..., a+(n-1)d

So n-1 = 18 and hence, n = 19

Using the formula given, you get
Sum = n/2(2a + (n-1)*d) = 19/2 * [2*26 + (19 -1)*2] = 836

So the required sum = 5050 - 836 = 4214

For more on arithmetic progressions, check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/03 ... gressions/

There are two parts to this question:

1st part: Sum of integers from 1 to 100 inclusive
We can find this by applying the formula
Sn = (n/2)*(2a + n-1)d.

The other way to write this formula is (n/2)*(a + l), where a = first term and l = last term

Applying the formula,
Sn = (100/2)(1+100) = 50*101 = 5050

Second Part: Sum of even integers between 25 and 63
The even integers start from 26 and end at 62
No of terms = 19

Hence the sum = (19/2)*(26 + 62) = 836

Subtracting this from 5050, we get 4214. Option D

Hi,
a method to do it under one minutes is to take advantage of the choices given...
lets work on the last digit as we have different units digit in each choice...

total sum of 1 to 100 inclusive will have 0 as the last digit..
this is so because the resultant will be 10*(sum of all single digits)... and since we are multiplying by 10, units digit will be 0...

now for single digit in sum of even number from 25 to 63..
25 to 65 will have 4 times sum of single digit even int, 4*(2+4+6+8+0)=4*20..
here too the units digit is 0, but 64 has to be excluded from the total..

two ways from here on..

1) we are subtracting 0 from 0
so units digit should be 0, but we have to add 64..
so last/units digit =4..

2)we subtract 64 from sum of even int..
so units digit=80-4=76..
or units digit =6...

so our answer should have units digit as 10-6=4..
only D has 4 as units digit..
ans D

We can find the sum of the integers 1 to 100 inclusive and subtract from it the sum of the even integers between 25 and 63 (i.e., the even integers from 26 to 62 inclusive).

Using the formula given in the question for the sum of the integers 1 to 100 inclusive, we let a = 1, d = 1, and n = 100; thus, the sum is:

100/2[2(1) + (100 - 1)(1)] = 50(101) = 5050

To calculate the sum of the even integers 26 to 62 inclusive, we let a = 26, d = 2, and n = (62 - 26)/2 + 1 = 19; thus, the sum is:

19/2[26(2) + (19 - 1)(2)] = 19/2(88) = 19(44) = 836

Therefore, the sum of the integers from 1 to 100 inclusive, with the sum of the even integers between 25 and 63 omitted, is:

5050 - 836 = 4214

Note: We don’t need to use the formula given in the problem; we can use the basic formula for the sum of an evenly spaced set of numbers, sum = quantity x average. For example, for the sum of the integers 1 to 100 inclusive, quantity = 100 and average = (1 + 100)/2 = 50.5. Thus, sum = 100 x 50.5 = 5050.

Answer: D

Can we apply sum = # terms x (first term + last term) / 2 for all sequences?

No, you can use for sequences which have equally spaced elements, that is they are in Arithmetic Progression.
For example
1,3,5,7
100,200,300,400
8,6,4,2

newdimension

What is your quick way to find that there is 19 int between 25 and 63?
for the sum 1 to 100< do we always add 1? for exemple, if it is sum from 3 to 97, we do 97+ 3/2 +1?

\(Number \ of \ multiples \ of \ x \ in \ the \ range = \)

\(=\frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).

Example 1: how many multiples of 5 are there between -7 and 35, not inclusive?

Last multiple of 5 IN the range is 30;
First multiple of 5 IN the range is -5;

\(\frac{30-(-5)}{5}+1=8\).

Example 2: How many multiples of 4 are there between 12 and 96, inclusive?
\(\frac{96-12}{4}+1=22\).

Example 3:
How many multiples of 7 are there between -28 and -1, not inclusive?
Last multiple of 7 IN the range is -7;
First multiple of 7 IN the range is -21;

\(\frac{-7-(-21)}{7}+1=3\).

So, the number of even numbers between 25 and 63 is (62 - 26)/2 + 1 = 19.

Next, the sum of the terms in any evenly spaced set (aka arithmetic sequence, arithmetic progression), which sets of consecutive integers belong to, is:
\(Sum = (\frac{first + last}{2}) * (number \ of \ terms)\), the mean multiplied by the number of terms;

So, the sum of the integers from 3 to 97, inclusive, is (3 + 97)/2 * 95 = 4,750.

Hope it helps.

Thank you very much !!!

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