Find n in arithmetic sequence
How can solve for n if n∈ℕ: $$ (4+8+12 + \ldots +4(n−1)) + (2+4.5+7 + \ldots +(4+5n)/2) = 126.5 $$Thank you
$\endgroup$ 31 Answer
$\begingroup$Hint:The first sum is:
$s_1=4(1+2+3+. . . n-1)=4\times\frac{(n-1)(n-1+1)}{2}=2n(n-1)$
The second sum $s_2$ is arithmetic progression $a=4.5$ and $d=2.5$ . We have:
$s_1+2+s_2=126.5$ ⇒ $s_1+s_2=124.5=3\times 41.5$
Let number of terms of $s_1$ be n and number of terms of $s_2$ be k, we can write:
$2n(n+1)+\frac{k}{2}[2\times 4.5 +(k-1)2.5]=3\times 41.5$
This relation deduces that n and k must be divisible by 3. Now we have to check multiples of 3 for n and k such as (n, k)=(3,3), (6,3), (3, 6) (6, 6) . . etc. We find that $n=k=6 $ is the correct answer, because:
$s_1=2\times 6(6-1)=60$
$s_2=3[2\times 4.5+(6-1)2.5]=64.5$
⇒ $s_1 + s_2=60+64.5=124.5$
$\endgroup$ 6
