Question

Given the sequence defined by $a_n = 48 - 3n$, find all possible values of $k$ so that \[a_1 + a_2 + a_3 + \cdots + a_k = 330.\]

Answer 1

k = 11 or 20

Explanation:

The sum of terms of an arithmetic sequence is given by ...

Sn = (a1 +an)(n/2)

We want to find the value of n for a particular series to have a particular sum.

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parameters

The series of interest is defined by ...

an = 48 -3n

so, the first term is ...

a1 = 48 -3(1) = 45

The k-th term is ...

ak = 48 -3k

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sum

Then the sum is ...

Sk = (45 +(48 -3k))(k/2)

We want this value to be 330, so we have ...

330 = (93 -3k)(k/2)

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solution

220 = (31 -k)k . . . . . . . . multiply by 2/3

k² -31k +220 = 0 . . . . . in standard form

We want factors of 220 that total 31.

220 = (1)(220) = (2)(110) = (4)(55) = (5)(44) = (10)(22) = (11)(20)

The last of these pairs totals 31, so we have ...

(k -11)(k -20) = 0 . . . . . factored form of the equation

k = 11 or k = 20 . . . . . values of k that make the factors zero

Possible values of k are 11 and 20.