Arithmetic Progression from Two Given Terms

An AP is fixed by two numbers: the first term $a$ and the common difference $d$. Give the examiner any two terms and you have two equations, subtract to eliminate $a$, find $d$, then back-substitute. The sum formula finishes the job.

In an arithmetic progression, the 5th term is $17$ and the 12th term is $45$. (i) Find the first term $a$ and the common difference $d$. [4] (ii) Find the sum of the first $20$ terms. [2]

The working

(i) Use $u_n = a + (n - 1)d$ for each given term: $u_5 = a + 4d = 17 \qquad u_{12} = a + 11d = 45 \quad \text{(M1 for both)}$

Subtract the first from the second (the $a$ cancels): $7d = 28 \;\Rightarrow\; d = 4 \quad \text{(A1)}$

Back-substitute into $a + 4d = 17$: $a + 16 = 17 \;\Rightarrow\; a = 1 \quad \text{(A1)}$

(ii) Use $S_n = \dfrac{n}{2}\big[2a + (n - 1)d\big]$ with $n = 20$, $a = 1$, $d = 4$: $S_{20} = \frac{20}{2}\big[2(1) + 19(4)\big] = 10\big[2 + 76\big] = 10 \times 78 = 780 \quad \text{(M1, A1)}$

Where the marks are won and lost

• Setting up both term equations correctly (the 5th term is $a + 4d$, not $a + 5d$, it is $a + (n-1)d$) is the first method mark.
• Subtracting to remove $a$ is cleaner than substitution. $u_{12} - u_5 = 7d$ links directly to $45 - 17 = 28$.
• In the sum, the bracket is $2a + (n-1)d$ with $n - 1 = 19$, not $20$. Using $20$ inside is the standard slip.

Common mistakes

• Using $a + 5d$ for the 5th term (off-by-one on $n - 1$).
• Putting $n = 20$ where $(n - 1) = 19$ belongs in the sum formula.
• Arithmetic slips in $10 \times 78$; keep the bracket evaluated before multiplying.

Full method: Arithmetic Progressions notes. Topic home: Series pillar.