Sum of an Arithmetic Series

A club needs six hundred and sixty-six pounds for new kit, so somebody suggests a plan. Three pounds into the tin this week, and four pounds more than that every week after. The payments run three, seven, eleven, fifteen, climbing by four every time. You could add them up week by week until the tin gets there, and that is a long column of additions with a slip waiting on every line. The question underneath is a different shape from anything a sequence rule answers. It is not what is week nine worth. It is how many weeks does this take. Check the specification you are sitting before you settle in. Totalling a series like this belongs to Edexcel International GCSE Higher, and the other GCSE maths papers do not ask for it.

Adding up a run of terms like that has a shortcut, and the shortcut is worth understanding rather than trusting. Take the first six weeks: three, seven, eleven, fifteen, nineteen, twenty-three. Pair the outside two and you get twenty-six. Pair the next two inwards, seven and nineteen, and it is twenty-six again. Every pair matches, because each step up one side is cancelled by an equal step down the other. Three pairs of twenty-six is seventy-eight, which totals six weeks of payments in one multiplication. In symbols that is the sum formula. S n equals n over two, times, two a plus n minus one d. The n over two counts your pairs, and the bracket is what one pair adds to. a is the first term, d is the common difference, and n is how many terms you are adding up. Quick check before we use it. Which of these is the correct substitution for ten weeks of the tin? A, ten over two, times, three plus nine fours. B, ten over two, times, two threes plus nine fours. C, ten over two, times, two threes plus ten fours. Take your pick. I'll wait. The answer is B. A forgot to double the first term, and C used ten steps of four when ten terms have only nine gaps between them. Five times forty-two is two hundred and ten pounds after ten weeks, nowhere near six hundred and sixty-six. Every interesting question here starts where that one stopped.

Turn the formula round now. You know the total you are aiming at, and what is missing is how many weeks it takes. Worded the way a paper words it: an arithmetic series has first term three and common difference four, the sum of the first n terms is six hundred and sixty-six, form and solve a quadratic in n. Everything but n is known, so put a as three and d as four in. That gives n over two, times, six plus four lots of n minus one, equals six hundred and sixty-six. Tidy the bracket first. Four lots of n minus one is four n minus four, and six minus four leaves four n plus two. Halve the bracket rather than the n, because four n plus two halves cleanly. That leaves n times two n plus one, so the left side is two n squared plus n. Bring everything to one side and there is your equation: two n squared plus n minus six hundred and sixty-six equals zero. Look at why a squared term turned up at all. In the formula n sits outside the bracket and inside it, so solving for n multiplies n by n. Two n's bend it into a quadratic. Solving quadratics is a topic of its own, owned by the video called Rearranging and Factorising a General Quadratic Equation. In brief, this one factorises into n minus eighteen, times, two n plus thirty-seven. Set each bracket to zero and two solutions drop out: n equals eighteen, and n equals minus eighteen point five. Both of those genuinely solve the equation, and only one can answer the question. Which one is it, and what is your reason? Have a think. I'll wait. The answer is n equals eighteen. Here n counts terms, so it has to be a positive whole number, and minus eighteen point five fails on both counts. There is no such thing as minus eighteen and a half payments. Write that reasoning down as a line of working, because the choosing is part of the answer. Then check it back through the formula. Eighteen over two is nine, and two threes plus seventeen fours is seventy-four. Nine times seventy-four is six hundred and sixty-six. Eighteen weeks and the club has its kit money. The algebra handed you two numbers, and the last line was reading the situation.

Not every series question leaves n as the missing piece, and the difference shows from the very first line. An arithmetic series has a fifth term of seventeen, and the sum of its first eight terms is one hundred and forty-eight. Find the first term a and the common difference d. Notice what is known this time. The number of terms is handed to you twice, five and eight, and a and d are the two things missing. The fifth term gives your first equation. A term of an arithmetic sequence is a plus n minus one d, so the fifth term is a plus four d, and that equals seventeen. Building single terms is the video called Finding the nth Term of a Linear Sequence. The sum of eight terms gives your second. S eight is eight over two, times, two a plus seven d, equals one hundred and forty-eight. Four times the bracket is that total, so the bracket itself is thirty-seven. Both came out straight, with no squared term anywhere. In the formula a appears once, d appears once, and the two are never multiplied together. Two n's bend it, one a and one d keep it straight. Solving a pair like that is its own topic, in the video called Solving Simultaneous Linear Equations by Elimination. In brief, the first gives a as seventeen minus four d, and putting that into the second leaves thirty-four minus d equals thirty-seven. So d is minus three and a is twenty-nine. This series starts at twenty-nine and steps down by three each time. Check both facts before moving on. Twenty-nine, twenty-six, twenty-three, twenty, seventeen gives the fifth term, and S eight comes out as four times thirty-seven. No quadratic in that one, and so no root to choose between. What you were given decided that before you wrote a single line.

You can make that diagnosis before writing anything down, and it is worth doing on a fresh question. A series has first term five and common difference six. The sum of the first n terms is one thousand two hundred and forty. Are you heading for a quadratic in n, or for two straight-line equations in a and d? Pick one. I'll wait. The answer is a quadratic in n. Both a and d are handed to you, so n is the only letter missing, and two n's bend it. So read a question for what is absent, not for what is there. A missing n gives one equation that bends, and a missing a and d give two that stay straight.

Now the last line of working, and the mark that hangs on whether you write it. This comes from a report on a Higher International GCSE paper sat in June twenty twenty-three, about an arithmetic series question on that paper. It was pleasing to see some students gain full marks on this arithmetic series question. Of those that did not, some were able to set up two linear equations in a and d but go no further. Those that found values for a and d then had difficulty solving the quadratic equation successfully. Some students made it all the way to solving a correct quadratic to gain answers of 57 and minus 24 but did not select 57 as their answer, losing the A mark. Three separate stopping points sit in that one paragraph. Equations set up and abandoned. Values found but the quadratic unfinished. And a quadratic solved with nothing chosen at the end. That last group had done everything difficult. They had two numbers in front of them, the question wanted one, and the report says the mark went for not naming it. So keep one sentence ready, written directly under your two solutions: n is a number of terms, so it must be a positive whole number, which makes n eighteen and rejects the other. Now finish the question you diagnosed a moment ago. First term five, common difference six, and a sum of one thousand two hundred and forty. Form the equation in n. Then, given that it factorises to n minus twenty, times, three n plus sixty-two, write the finishing line. Pause it there and work it out. I'll wait. Substituting gives n over two, times, ten plus six lots of n minus one. The bracket tidies to six n plus four, and halving it leaves n times three n plus two, or three n squared plus two n. Set that equal to one thousand two hundred and forty and the brackets give n equals twenty, or n equals minus sixty-two over three. n counts terms, so n is twenty and the negative fraction is rejected. Twenty terms, with the reason for choosing twenty written where a marker can see it.

Wind it back, because this whole topic is one formula and one decision. The sum of the first n terms is n over two, times, two a plus n minus one d, and it works because pairing from the outside inwards gives equal-sized pairs. Then read the question for what is missing. Two n's bend it, one a and one d keep it straight. A missing n means substituting a and d in, tidying the bracket, and rearranging into a quadratic that equals zero. A missing a and d mean two straight-line equations instead, one built from a term and one from a sum, then solved together. And a quadratic in n always ends in a choice, because n counts terms. A negative or a fraction gets rejected on the page, not just in your head. You have got this when two roots from a series question leave you able to say, in one sentence, why only one can be the number of terms.

That completes our chapter on Sequences and Series. The next chapter is Functions and Differentiation.

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