[mail2abhishek99]

Find the sum:
1+4+7+10+... to 100 terms

[Alty]

What answer did you get? Show us your work, and your answer and we'll go from there.

We won't do your homework for you, it's your assignment and you have to do the work. We can offer advice on your work, or tell you where you went wrong, but we won't give you the answer. That's up to you.

[Unknown008]

You need to first identify the pattern, whether it's an arithmetic progression or a geometric progression or some other sequence. If it's the first case, you will need to use the formula:



If it's the second, then you'll need to use:



If it's the third, you'll have to work out every term in the pattern and add them up, unless you come up with a formula yourself.

In any case,
a is your first term,
n is the nth term (in your case, 1 gives n=1, 4 gives n=2, 7 gives n=3, 10 gives n=4, etc. You'll have to find what 'rank' is 100),
d is the common difference,
r is the common ratio.
Sn is finally the sum of all the terms, until the nth term

Sn for n=1 (that is for the first term) is 1
Sn for n=2 gives 1+4 = 5
Sn for n=3 gives 1+4+7 = 12

etc.

Post what you get! :)

[Alexey_u80]

This is arithmetic progression. D = 3 a(1) = 1, and a(100) = a(1)+d * (100 - 1) = 1+3*99= 298. The sum of the first n terms is S(n) = (a(1)+a(n))/2*n = (1 + 298)/2*100 = 14950

[Unknown008]

No, that's wrong Alexey, 100 is not the 100th term.

[Alexey_u80]

100 is the number of the last elemet of the sum.

[Alexey_u80]

a(1)+a(2)+a(3)+... +a(100). a(100) is 298 in my account.

[Unknown008]

Hm... I see now what you mean, I initially took 100 as being the last term while you took it as being the 100th term.

Anyway, that said, you are not supposed to do the homework of the members asking questions. It's better to show them how to do the question, give them options like I did and let them work it out, and post what they get.

One more thing, it's the first time that I see that notation, that is using a to describe the term. I learn it using T, T1 being the first term (or in the formula I gave, as a), then T2, T3, T4, etc. :p

[jcaron2]

Since Unknown and Alexey gave you the formula (and the final answer in Alexey's case), I'll at least give you a quick explanation of how you arrive at that formula:

The first number is 1, and the 100th number is 298. Those two numbers add up to 299.
The second number is 4, and the 99th number is 295. Those two numbers also add up to 299.
In fact, if you keep pairing them together, 1st + 100th, 2nd + 99th, 3rd + 98th,. 50th + 51st, you'll see that there are 50 pairs of numbers, all of which add up to 299. Hence, the sum is 299*50 = 14950, just like Alexey said.

[Alexey_u80]

Jcaron2 gave very helpful. So we may extent it in common case. Let us plus two sums: S(n) + S(n) = 2S(n). We will write it one from the left to wright and other from the wright to left : 2*S(n) = a(1) + a(n) + a(2) + a(n-1) +.. +a(n)+a(1), and a(1) + a(n) = 2 * a(1) + d, a(2) + a(n-1) = 2* a(1) + d and so on. The number of such small sums is n, and 2 S(n) = (2* a(1) + d)*n, and S(n) = (2* a(1) + d)* (n/2), but a(1)+d = a(n). And S(n) = (a(1) + a(n)) * (n/2). That is the way which one can get it well-known formula.