Sum of first n terms of an AP

An arithmetic progression (AP) is a list of numbers in which each term differs from the one before it by a fixed amount, the common difference $d$. Denote the first term $a$, and let $S_n$ stand for the sum of its first $n$ terms — not any single term, but the running total. For any AP with first term $a$ and common difference $d$, the sum of the first $n$ terms is

where: $a$ is the first term, $d$ is the common difference, $n$ is the number of terms being added, and $S_n$ is their sum. The bracketed quantity $2a+(n-1)d$ is exactly twice the average of the first and last term — the formula in disguise. If the last term $a_n$ (sometimes written $l$) is already known, a shorter equivalent form avoids recomputing $d$ altogether:

where: $a$ is the first term and $a_n$ is the $n$th (last) term being summed. This second form reads simply as "number of terms times the average of the first and last term," and it is the more convenient choice whenever a problem hands you the last term directly. The two forms are not independent facts to memorise separately — the second collapses into the first the instant you substitute $a_n = a+(n-1)d$. Reach for $S_n=\tfrac{n}{2}[2a+(n-1)d]$ when you know $a$, $d$ and $n$ but not the last term; reach for $S_n=\tfrac{n}{2}(a+a_n)$ when the last term is already sitting in front of you. A companion idea, tested just as often, links a sum back to a single term. Since $S_n$ is the running total through the $n$th term while $S_{n-1}$ stops one term earlier, their difference isolates that one term:

where: $S_n$ is the sum of the first $n$ terms and $S_{n-1}$ is the sum of the first $n-1$ terms, so their difference is whatever got added last, namely $a_n$. This relation is useful even when the AP's rule is not given explicitly, as long as a formula for $S_n$ in terms of $n$ is known. The picture below shows why pairing works: stack the terms as bars of increasing height and lay a second, upside-down copy of the same staircase against it — together they tile a rectangle of constant height $2a+(n-1)d$ and width $n$, so twice the sum is that rectangle's area.

Question. Find the sum of the first 20 terms of the AP $2, 5, 8, 11, \ldots$, and then use that sum together with $S_{19}$ to recover the 20th term directly.

Step 1. Identify $a$ and $d$. Here $a = 2$ and $d = 5-2 = 3$. We want $S_{20}$, so $n=20$.

Step 2. Apply $S_n = \dfrac{n}{2}[2a+(n-1)d]$: $S_{20} = \dfrac{20}{2}\big[2(2)+(20-1)(3)\big] = 10\big[4+57\big] = 10\times 61 = 610$. So the sum of the first 20 terms is $\boxed{610}$.

Step 3. Cross-check with the other form. The 20th term is $a_{20} = a+19d = 2+19(3) = 59$, so $S_{20} = \dfrac{20}{2}(a+a_{20}) = 10(2+59) = 10\times 61 = 610$ — the same answer, confirming both forms agree.

Step 4. Recover $a_{20}$ from two sums, without ever using the $n$th-term formula. Compute $S_{19} = \dfrac{19}{2}\big[2(2)+(18)(3)\big] = \dfrac{19}{2}\big[4+54\big] = \dfrac{19}{2}(58) = 19\times 29 = 551$. Then $a_{20} = S_{20}-S_{19} = 610-551 = \mathbf{59}$, exactly matching Step 3 and illustrating $a_n = S_n - S_{n-1}$ in action.

Competitive stretch

A problem may instead give you the sum and ask for $n$. For the AP $9,17,25,\ldots$ (here $a=9$, $d=8$), asking how many terms give $S_n=636$ leads to $636=\dfrac{n}{2}[18+8(n-1)]$, which simplifies to $4n^2+5n-636=0$. Factoring gives $(4n+53)(n-12)=0$, so $n=12$ or the rejected negative root $n=-\tfrac{53}{4}$. Check: $S_{12}=\dfrac{12}{2}[18+11(8)]=6(106)=636$. ✓

• Mixing up $a$ and $a_n$. The first form uses $a$ (first term) inside $2a+(n-1)d$; the second form uses $a_n$ (last term) directly. Do not plug the last term into the slot meant for the first term.
• Forgetting the $\tfrac{n}{2}$ factor. Computing $2a+(n-1)d$ correctly and then forgetting to multiply by $\tfrac{n}{2}$ is one of the most common arithmetic slips.
• Keeping the wrong root of the quadratic. When solving for $n$ from a given $S_n$, always discard negative or non-integer roots, because $n$ counts terms and must be a positive whole number.
• Confusing $S_n$ with $a_n$. $S_n$ is a cumulative total; $a_n$ is a single term. Using a sum formula where a term formula is needed (or the reverse) leads to the wrong equation entirely.

Ask the tutor: If a quadratic pops out while solving for $n$ and you are unsure which root to keep, ask the AI tutor to walk through the factoring and the reasoning for rejecting the invalid root.