Math · Precalculus · Worked example

Find a term and a sum of an arithmetic sequence

For the arithmetic sequence 5, 8, 11, 14, …, find the 20th term and the sum of the first 20 terms.

Find the common difference

Each term is 3 more than the one before, so a₁ = 5 and d = 3.

8 - 5 = 11 - 8 = 14 - 11 = 3

Write the nth term

Substitute a₁ = 5 and d = 3 into aₙ = a₁ + (n − 1)d, then simplify.

\begin{array}{l} a_{n} = 5 + 3 (n - 1) \\ a_{n} = 3 n + 2 \end{array}

Find the 20th term

Put n = 20.

3 \cdot 20 + 2 = 62

Add the first 20 terms

The sum is the number of terms times the average of the first and last terms.

\frac{20 (5 + 62)}{2} = 670

Result

a₂₀ = 62, and the first 20 terms add to 670.

Your turn

Find the 15th term and the sum of the first 15 terms of 40, 37, 34, ….

Show the answer and explanation

a₁₅ = −2 and S₁₅ = 285.

d = −3, so a₁₅ = 40 + 14(−3) = −2, and S₁₅ = 15(40 + (−2))/2 = 285.

\begin{array}{l} 40 + 14 \cdot (- 3) = - 2 \\ \frac{15 (40 + (- 2))}{2} = 285 \end{array}

Keep exploring

In Sequences & infinite series, the table lists all 20 terms and their running totals, ending at 62 and 670. Change the count to 50: the last term is 152 and the total 3925.

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