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## Class 10 (Old)

### Course: Class 10 (Old) > Unit 5

Lesson 4: Sum of first n terms of an AP- Arithmetic series intro
- Arithmetic series formula
- Worked example: arithmetic series (sum expression)
- Finding first term and common difference when sum is given
- Finding number of terms when sum of an arithmetic progression is given
- Sum of n terms (intermediate)
- Sum of n terms (advanced)
- Comparing arithmetic progressions

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# Worked example: arithmetic series (sum expression)

Sal evaluates the arithmetic sum (-50)+(-44)+(-38)+...+2044. He does that by finding the number of terms and using the arithmetic series formula (a₁+aₙ)*n/2.

## Want to join the conversation?

- At4:36, Sal says "-50 + 2044 is equal to 2094"; however, when I added 2044 to -50 by hand and on a calculator the answer I got was that -50 + 2044 = 1994. Did I make a mistake in my math?(7 votes)
- at5:07Sal corrects his mistake by saying, "my brain really wasn't working a while ago"(6 votes)

- I got n to be 348. Did I make a mistake in my algebra?(5 votes)
- You may have subtracted 1 instead of adding 1 in the last step.

The nth term of an arithmetic series is given by a_n = a + d(n - 1), where a is the first term and d is the common difference.

We know that a = -50, d = 6, and a_n = 2,044.

So 2,044 = -50 + 6(n - 1); 2,094 = 6(n - 1); 349 = n - 1; n = 350.(23 votes)

- Wouldn't it be easier to 0 index all these series instead of 1 indexing them? That way we don't need to put k-1 but can instead put k?(5 votes)
- That's what I did as well. I don't think it makes any difference, but it is important to note that if you do it this way, then your Summation will go from k=0 to 349 instead of k=1 to 350.(10 votes)

- I'm very confuzzled on this can someone like explain it better?(4 votes)
- Could you please specify at what time in the video, so that I don't have to watch the full video to figure out where would be your question regard to.(2 votes)

- Instead of using Sal's method to get 350 terms, I found the formula next to the sigma symbol, which was 6k - 56. Then, I put this equal to 2044 and I also got 350 terms. Is this also a viable method, or did I just get lucky?(5 votes)
- That would also be a working method to find out how many terms there are. What you essentially did was find the explicit formula for the sequence and used that to find the term at which the sequence would equal 2044.(2 votes)

- can I use this formula for

3 + 16 + 29 +42...+2044

I figured n=158. My answer was 161,713. Is this correct?(3 votes)- As long as it is an arithmetic sequence, it is okay to use the formula.(3 votes)

- Why-did-he-divide-350-by-2(2 votes)
- because part of the formula is to divide the sum of the 1st term + the last term by 2(2 votes)

- Why can't the rule be written as -56+6k instead of -50+6(k-1)?(1 vote)
- Your version is equivalent. It is just a simplified form of what's in the video. Either will work.(4 votes)

- how do you find the equation on the right of the sigma if you are given the sequence and the partial sum?(2 votes)
- In the sigma equation, why does he set
**K=1**? Does**K**has another option?(2 votes)

## Video transcript

- [Voiceover] So we
have the sum negative 50 plus negative 44, plus
negative 38, all the way, we keep adding all the way
up to 2,038, and then 2,044. So see if you can pause this
video and evaluate this sum. So let's work through this together, and let's just think
about what's going on. So the first term here is negative 50, and then we go to negative 44, so the second term is negative 50 plus 6, and then the third
term, we add 6 again. Negative 44 plus 6 is negative 38, and we go all the way to here, we keep adding 6, and to go from 2,038 to 2,044,
to get to that last term, we add 6 once again. And so each successive term is just 6 more than the term before it. So we are dealing right over here, this sum is an arithmetic series. It's a sum of an arithmetic sequence. Each term is 6 more, is a constant amount more than the term before that. So we know how to take the sum of an arithmetic sequence. We know that if we have, if we are taking the sum of, let me do this in a new color, just to have a little bit
of variety on the screen, if we're taking the sum of, of the first n terms of
an arithmetic sequence or, if we're taking or if we're evaluating the
first n terms of an arithmetic series, I could say, it's going to be the first term plus the last term divided by 2. You could view this of
the average of the first and the last terms, times the number of terms that we're dealing with. So over here, we know what
our first and last terms are, we know this right over here. That is a1, and this is our last term, 2,044, so that is our a-sub-n, so the other question is, well, what is n? How many terms do we actually deal with? And to think about that, we just say, well how many times do we have to add 6 to go from negative 50 to 2,044? Well, 2,044 minus negative 50, minus negative 50, well that's the same
thing as 2,044 plus 50 or 2,094, and the whole
reason I calculated this is I want to figure out
how far do I have to go from negative 50 to 2,044? I have to go up 50 just
to get back to zero and then go up another 2,044. So I have to go 50 just
to get back to zero, then go up another 2,044 so I have to go up by 2,094, so if I'm going, if I'm adding 6 on every term, how many times do I have to add 6 to increase by 2,094? Well, let's just take
2,094 and divide it by 6 to figure that out. So 6 goes into 20 three times, 3 times 6 is 18, subtract, 20 minus 18 is 2, bring down the 9, 6 goes into 29 four times, 4 times 6 is 24, subtract, 29 minus 24 is 5, bring down the 4, we have 54, 6 goes into 54 nine times, 9 times 6 is 54 and we are done. So to go from negative 50 to 2,044, I have to add 6 to 349 times, so I add it once, I add it twice, and then this right over here, this is the 349th time that I'm adding 6, so how many terms do I have? Now, you might be
tempted to say 349 terms, but really, you have 349 plus 1 terms. You have the 349 for
every time you added 6, so this is the first time you added 6, second time you added 6, all the way to the 349th time you added 6, so let me make it clear, this, this is, oh, actually, this is the
349th time I added 6 to get to this, but we haven't counted
the first term just yet, so we're going to have, so we have 300, we have the first term and
then we add 6 349 times, so we have 350 terms in this sum, so in this case, n is
going to be equal to 350. N is equal to 350. And so we can say the
sum of the first 300... I'll do this in green, the sum of the first 350 terms is going to be equal to the
average of the first and last term, so negative 50 plus 2,044 over 2, over 2, times 350, so let's see, negative 50 plus 2,044, that's going to be what? 2,094, 2,094 divided by 2 times, times 350, so let's see, if I just take, so this is going to be, 'cause
right, this is 294 times, let's see, 350 divided, oh, sorry, not 294, what am I? My brain is not working. 2,000, this is 2-- actually, this is going to be 1994. My brain really wasn't
working a little while ago, so this is going to be
1994 divided by 2 times 350 and so let's see, 350 divided by 2 is 175, so this is going to be 1,900-- 1,994 times 175. Which is equal to, and I'll use a calculator for this one, so, let me get the calculator out, so, I have 1,994 times 175 gives us 348,950. 348,950. And we could express this
in sigma notation now, now that we know what the n is. We found our answer, this
is what we were looking for, but just in case you're curious, we could write this as the sum from, let's say, k is equal to 1 to k is equal to 350 of, let's see, we could write this as, as negative 50, negative 50 plus 6 times k minus 1, 'cause the first term,
we don't want to actually add the 6, and then the last term, we want to add the 6 349 times. Which we saw, so we're
going to add it 349 times, and there you have it. That, this arithmetic series
written in sigma notation, so hopefully you enjoyed that. Pardon my little mental error earlier. I don't know what was
going on in my brain.