Arithmetic Sequences IB Math AA SL: Paper 1 No-Calculator Question, Step by Step [IB1]

Arithmetic sequences, for I-B Math A-A S-L. This answer scores zero.

The student added ten fours to the first term. Can you see the mistake?

To get from the first term to the tenth, you take nine steps, not ten. Three plus nine fours. Thirty-nine.

That's one mark of a six-mark Paper 1 question, with no calculator. Five more marks, and each one hides a trap like that.

So why nine steps? And how do you write any term in one go?

Look at the terms. Three, seven, eleven, fifteen. The first term has no fours added. The second has one four. The third has two. So term n has n minus one fours.

That's the formula in your data booklet. u n equals u one, plus n minus one, times d. Here, three plus n minus one, times four.

Now you simplify it. Is it four n plus three, four n minus one, or four n minus four?

Four n minus one. Four n, minus four, plus three. Check it with n equals one. Four minus one is three, the first term.

But part b says show that. In a show that question, the answer is already printed. The mark is for the working. So write the line with the brackets, then expand it. Just testing n equals one scores nothing. Mark two. So what do the first twenty terms add up to?

The data booklet gives you two sum formulas. S n equals n over two, times two u one, plus n minus one d. Or, if you know the last term, n over two, times first plus last.

The second one is the quick one. Here's why it works. The last term is u twenty. Four times twenty, minus one. Seventy-nine. Pair it with the first term. Three plus seventy-nine. Then seven plus seventy-five. Every pair makes the same number. And twenty terms make ten pairs.

You're the examiner now. Three students wrote the sum. Give each one zero, one or two marks.

Answer A is seventy-nine. That's the last term, not the sum. Zero. Answer B has the right formula. But six plus seventy-six is eighty-two, not eighty. One mark for the method. Answer C has ten pairs of eighty-two. Eight hundred and twenty. Two marks.

With no calculator, a slip like B's is the real danger. So write the bracket out before you multiply. And check that it makes sense. The average of the first and last terms is forty-one. Twenty terms of forty-one. Eight hundred and twenty. That's marks three and four. Now the last part turns the question round.

Which term of the sequence equals one hundred and twenty-three? This time you know the term, and n is the unknown. Pause if you want to try it first; here it is.

Use your formula from part b. Four n minus one equals one hundred and twenty-three. Add one. Four n equals one hundred and twenty-four. Divide by four. n equals thirty-one. Marks five and six.

And n has to be a whole number. If you'd got thirty point five, one hundred and twenty-three wouldn't be in the sequence at all.

How long should this question take you? Paper 1 at S-L is eighty marks in ninety minutes. That's a little over a minute per mark. So six marks is about seven minutes. Use the spare minute to check. Four times thirty-one, minus one. One hundred and twenty-three.

Now the same idea, backwards. Another question tells you u three is eleven, and u seven is twenty-seven. How many steps from the third term to the seventh? Three, four, or five?

Four. Seven minus three. You count the gaps, not the terms, just like at the start. Twenty-seven minus eleven is sixteen. Sixteen over four. d equals four. And u one is eleven minus two fours. Three. It's the same sequence.

Steps from u one to u ten? Nine. The nth term? u one plus n minus one d. The sum? n over two, first plus last.

Can you read a graph you've never seen before? Next Saturday, a Biology data question.