Given the sequence defined by a_{n}=48-3n, find all possible values of k so that

\(a_1 + a_2 + a_3 + \cdots + a_k = 330.\)

If you find more than one, then list the values separated by commas.

Guest May 14, 2022

#1**0 **

Note that a_n is just an arithmetic sequence. You can use the sum of A.S. formula to rewrite the equation a_1 + a_2 + ... + a_k = 330.

MaxWong May 14, 2022

#3**0 **

You know that

\(a_1 = 48 - 3\times1 = 45\\ d = a_2 - a_1 = (48 - 3(2)) - (48 - 3(1)) = -3\\ \)

where d is the common difference of the sequence.

The sum of A.S. formula states \(a_1 + a_2 + \cdots + a_n = \dfrac n2 \left(2a_1 + (n - 1)(a_2 - a_1)\right) = \dfrac n2 (2a_1 + (n - 1)d)\) if a_n is an arithmetic sequence. Therefore, the equation \(a_1 + a_2 + a_3 + \cdots + a_k = 330\) really says \(\dfrac k2 (2(45) + (k - 1)(-3)) = 330\). This becomes a quadratic equation. Can you solve for the value of k now?

Hint: Expand and use quadratic formula.

MaxWong
May 15, 2022