How can I find the first n terms in an arithmetic sequence if the last term isn't given

Question

I have two questions.

1. Suppose I have an arithmetic sequence $11,20,29,38,...$ and $11+20+29+38+... = 8998$. How can I find the first $n$ terms that would get me to 8988?

I know that the sum is $\frac{n(a_1+a_n)}{2}$ and I have $8988 = \frac{n(11+a_n)}{2}$ or $17996 = n(11+a_n)$

Since I have two unknowns, I don't know how to solve this and I don't know how the formula for finding the $a_n$th term would help.

This is my second question:

2. Find all integers $N$ such that

$46+44+42+...+N = 510$

My main confusion is that I would need to find different values of $n$ which are the number of terms to then find different values of $N$ which would lead to 510.

So I know the common difference is 2, the number of terms is unknown, the first term is 46, the last term is unknown but the sum is 510.

So I end up in the same spot:

$1020 = n(46+a_n)$ Should I divide by $n$ and subtract 46 and get $\frac{1020}{n} - 46 = a_n$ and then just keep plugging in numbers?

Any help or hints are appreciated bc I'm really stuck, thanks!

Answer 1

But $a_{k+1} = a_k + 9$ so $a_n = 11 + (n-1)*9$.

So $a_1 + ...... a_k = \frac {n(a_1+a_n)}{2}= \frac {n(a_1 + a_1 + (n-1)*9}2= \frac {n(2*11 + (n-1)*9)}2 = =8998$

Solve for $n$ and....

2) Let $N = 46 - 2*(n-1)$ so then $46 + 44 + 42 + ..... +N = \frac {n(46 + N)}2 = \frac {n(46 + 46 - 2*(n-1)}2 = 510$

Answer 2

Given the sequence 11,20,29,38, first find a formula for $a_n$. Then worry about the sum.

Answer 3

HINT

for the second part, your sequence $46, 44, 42, \ldots$ is given by $a_k = 46-2k$ for $k = 0, 1, \ldots$ so you have a sequences of length $n$, and its sum is given by $$ s_n = \sum_{k=0}^n a_k = \sum_{k=0}^n (46-2k) = 46(n+1) - 2 \sum_{k=0}^n k = \ldots $$ Can you complete evaluating this in iterms of $n$ only? And then solve the obvious equation $s_n = 510$ for $n$?

Answer 4

$1.$ – This sequence has $9$ as a common difference, so the $n$th term is $\;a_n=2+9n$. Therefor the sum up to the $n$th term is equal to $$\frac{n(11+a_n)}2=\frac{n(13+9n)}2=8998\iff 9n^2+13n-17996=0.$$ Now we have to find a positive integer, so by the rational roots theorem, it should be a divisor of $$17996=2^2\,11\,409.$$ You can check $n=44$ is the solution, so that $a_n=398$.

$2.$ – Same method: we have an arithmetic progression with common, difference $-2$, so the $n$th term is $b_n=48-2n$ and the sum up to this term is equal to $$\frac{n(94-2a_n)}2=510\iff n^2-47 n+510=0$$ The possible value is a divisor of $510=2\,3\,5\,17$. As $n(47-n)=510$, looking at a rough estimate of the l.h.s., you can eliminate some values without really computing.