I have two questions.
- 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:
- 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!
Best Answer
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 $46 + 44 + 42 + ..... +N = \frac {n(46 + N)}2 = \frac {n(46 + 46 - 2*(n-1)}2 = 510$