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## Comment on

Sum of 40 to 60## If it was not inclusive then

## The sum of integers from 41

The sum of integers from 41 to 59 inclusive is 950.

Using the same strategy as in the video, there are 19 integers from 41 to 59 inclusive.

So, if we add the values in PAIRS again, we get 19 100's.

19 x 100 = 1900, and then we divide by 2 to get 950

## I just came across a question

## Be careful, that question is

Be careful, that question is asking for the sum of the EVEN integers from 100 to 300 inclusive.

Here's the question (and my solution): https://gmatclub.com/forum/for-any-positive-integer-n-the-sum-of-the-fir...

Cheers,

Brent

## Hi Brent,

I have a question regarding the above-mentioned video: You included 1 to 60 but why you have not included 1 to 40... you just calculated 1 to 39 only.

I have understood the concept but not this particular detail.

Thank you for your prompt reply

Fatima-Zahra

## We want to find the sum of 40

We want to find the sum of 40 to 60 inclusive.

So, once we find the sum of 1 to 60 inclusive, we must subtract all of the values that are NOT part of the sum from 40 to 60 inclusive.

So, we must subtract a 1 and a 2 and a 3 and . . . . . and a 39.

This will leave us with the sum of 40 to 60 inclusive.

Does that help?2

## Thank you Brent for the

Building on your explanation, can we derive a generic equation for all cases of sequence summation as following:

Sum of Sequence =(y+x)*(y-x+1)/2

where y is the last number and x is the first number in the sequence.

Thank you,

Ahmed

## Good idea!

Good idea!

In fact, that formula can be used to find the sum of CONSECUTIVE integers from x to y.

However, it won't work for other sequences.

Take, for example, the sequence 15, 20, 25

Here, x = 15 and y = 25

From your formula, the sum = (25 + 15)(25 - 15 + 1)/2 = 220

However, the sum is actually 60

Cheers,

Brent

## Hey Mr. Brent,

Simply put, n= the number of terms in the sequence?

Can I interpret it this way?

Thank you:)

## Great question!!

Great question!!

n indicates WHERE a term is located in the sequences.

For example, in MOST cases, term_3 is the 3rd term in the sequence, and term_11 is the 11th term in the sequence, etc.

The reason for this is that MOST questions start by saying that term_1 is the first term.

Here's an example (https://gmatclub.com/forum/a-sequence-of-numbers-a1-a2-a3-is-defined-as-...) in which the terms are defined as a1, a2, a3, ... etc.

HOWEVER, there are times when the test-maker lets the first term be something like a0.

In this example (https://gmatclub.com/forum/in-the-sequence-x0-x1-x2-xn-each-term-from-x1...), the terms are defined as x0, x1, x2, x3,... (very rude :-)

In these cases, we must be careful with our term numbers, since the third term is x2, and the fourth term is x3, etc)

Cheers,

Brent

## GOT MANY OF THESE QUESTIONS

I usually thought they are pretty hard but now I think I TOTALLY get them, using two approaches I can do literally anything!

## That's fantastic!

That's fantastic!

## Hi Brent,

I have really been enjoying your course so far. I am curious about the question linked below this video but the solution seems to be no longer available. I got C as the solution but I'd love to be able to check if this is correct and if not, how to solve it.

Thank you!

## I'm glad to hear you're

I'm glad to hear you're enjoying the course. You're right; the linked question isn't showing any solutions anymore. So, I deleted the link.

Here's the question (it was hard to tell by the formatting how statement 2 is supposed to look, but I think I have it).

The first term in a sequence is t, and each term thereafter is k/11 greater than the term before it. If t and k are both positive integers but only one of them is a prime number, what is the sum of the first 100 terms in the sequence?

(1) 2t + 11k = 46

(2) 3k = 14 - (2/3)t

Let's list some terms:

term1 = t

term2 = t + k/11

term3 = t + 2k/11

term4 = t + 3k/11

.

.

.

term99 = t + 98k/11

term100 = t + 99k/11

So, the sum of the first 100 terms = 100t + (k/11)(1+2+3+4+...+98+99)

Simplify to get: the sum = 100t + (k/11)(4950)

Rephrased target question: What is the value of 100t + (k/11)(4950)?

(1) 2t + 11k = 46

This statement is actually sufficient because the question tells us that t and k are both positive integers but only one of them is a prime number.

When we restrict t and k two positive integers, there are only two possible solutions to the equation.

case i) k = 2 and t = 12

case ii) k = 4 and t = 1

Since only one of these solutions satisfies the condition that only one value is prime, it must be the case that k = 2 and t = 12.

Now that we know the values of k and t, we can answer the rephrase target question with certainty.

So, statement 1 is sufficient.

(2) 3k = 14 - (2/3)t

Multiply both sides of the equation by 3 to get: 9k = 42 - 2t

Rearrange: 9k + 2t = 42

Once again, since t and k are both positive integers, there are only two possible solutions

case i) k = 2 and t = 12

case ii) k = 4 and t = 3

In this case, both solutions satisfy the condition that only one value is prime.

If k = 2 and t = 12, then the sum = 100(2) + (12/11)(4950)

If k = 4 and t = 3, then the sum = 100(4) + (3/11)(4950)

Since the two possible solutions yield different answers to the rephrased target question, statement 2 is not sufficient.

Answer: A

## Is there an easier way to

## For this particular question,

For this particular question, once we see the variables are restricted to positive integers, we can start testing some possible values.

Let's do this with the first equation: 2t + 11k = 46

I'm going to focus on the k-values since k has the bigger coefficient (11), which means the value of 11k will increase a lot faster than the value of 2t will.

Here's what I mean:

The smallest possible k-value is 1.

Plug this value into the equation to get: 2t + 11(1) = 46

Rearrange to get: 2t = 35.

Since t must be an integer, there's no solution to this equation.

When k = 2, we get 2t + 11(2) = 46, and when we solve this, we get t = 12. That's one solution.

When k = 3, we get 2t + 11(3) = 46, which doesn't have an integer solution for t.

When k = 4, we get 2t + 11(4) = 46, and when we solve this, we get t = 1. That's another solution.

At this point, we can see that when k > 4, the value of t will always be negative.

So, we can be certain we've tested all possible values of k.

Does that help?

## It does! thank you