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Add, or multiply?

◆FRAME

Two lists. Look at how each one grows.

LIST ONE: 3, 7, 11, 15, 19. From each number to the next, what happened? $7 - 3 = 4$. And $11 - 7 = 4$. And $15 - 11 = 4$. The same amount is ADDED at every step.

LIST TWO: 2, 6, 18, 54, 162. Subtracting here does not settle: $6 - 2 = 4$, but $18 - 6 = 12$. So divide instead. $6/2 = 3$. And $18/6 = 3$. And $54/18 = 3$. The same amount is MULTIPLIED at every step.

Both are SEQUENCES — ordered lists built by a rule. Add up a sequence’s terms and you have a SERIES, which is a single number rather than a list.

The two rules pull apart fast. Both lists start small, but by the tenth term list one has reached 39 while list two is past 39,000.

So the question to ask of any sequence is this: add, or multiply?

Answer ADD and it is an ARITHMETIC progression, with a common difference. Answer MULTIPLY and it is a GEOMETRIC progression, with a common ratio. Every formula in this chapter belongs to one side or the other.

Test both. Subtract consecutive terms; if that is not constant, divide them instead. One of the two usually settles it.

Both of the frame’s lists on one scale: LIST TWO starts under LIST ONE and ends far above it.
The same ruler under both lists: equal arcs mean an A.P., doubling arcs mean a G.P.

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A sequence lists, a series adds

●KEY-TERM

Two words that get used interchangeably in conversation and must not be here.

A SEQUENCE is an ordered list of numbers, written $a_1, a_2, a_3, \ldots$ and generated by a rule. The rule is usually given as a formula for the nth term, $a_n$, so any term can be produced without walking through all the ones before it.

A SERIES is the SUM of a sequence’s terms, $a_1 + a_2 + \ldots + a_n$, written $S_n$.

So a sequence LISTS and a series ADDS. The sequence 3, 7, 11 is three numbers; the corresponding series $S_3$ is the single number 21.

Order is part of what a sequence is. Rearranged, it is a different sequence even with the same numbers in it, which is why the terms are subscripted by position.

That subscript notation carries the position everywhere in this chapter. $a_1$ is the first term, $a_n$ is the nth, and $S_n$ is the sum of the first n of them. Keeping those three apart is most of what is needed to read the formulas.

*A sequence is a list; a series is a total. $a_n$ is one term; $S_n$ is n terms added.*

Two rails on the same five numbers: the sequence itself above, the running total below.

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Arithmetic progressions

▲CONCEPT

An ARITHMETIC PROGRESSION, usually shortened to A.P., is a sequence where each term after the first is the previous one plus a fixed amount.

That fixed amount is the COMMON DIFFERENCE, written d. Starting from a first term a, the sequence runs $a$, $a+d$, $a+2d$, $a+3d$, and so on.

The nth term is $a_n = a + (n-1) d$.

The $(n-1)$ matters, so count steps rather than terms. The first term needed no steps of d at all. The second needed one. The third needed two. So the nth needed $n-1$ of them.

Check it at the start. Put $n = 1$ and you get $a_1 = a + 0 \times d = a$, which is the first term. Correct. Put $n = 2$ and you get $a + d$, also correct.

Two other things follow from the definition. The common difference d may be negative, in which case the sequence decreases. And because the same amount is added each time, the terms are evenly spaced — which is the property the sum formula will lean on.

*Test $n = 1$ on any formula of this shape. If it does not return the first term, the offset is wrong.*

Six terms on a number line with the five steps between them drawn as arcs underneath.
Worked example

Find the 10th term of 3, 7, 11, 15, ...

  1. $a = 3$; $7 - 3 = 4$, $11 - 7 = 4$, $15 - 11 = 4$
    Get the two constants first, subtracting consecutive terms to confirm d is constant — a list that is not an A.P. will not announce itself.
  2. $a_{10} = 3 + (10-1) \times 4 = 3 + 9 \times 4 = 3 + 36 = 39$
    Substitute into $a_n = a + (n-1) d$ with $n = 10$.
  3. nine steps of 4, not ten
    Getting to the 10th term takes nine steps of 4 — using ten would give 43, the 11th term, a wrong answer that looks entirely reasonable on the page.
  4. 3, 7, 11, 15, 19, 23, 27, 31, 35, 39
    Check it cheaply by extending the list — the tenth entry is 39.
Ten terms and their step counts side by side; the ringed column is the one the formula computes.

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Adding up an A.P.

▲CONCEPT

Adding the first n terms of an A.P. has a formula, and one worth deriving rather than memorising.

$S_n = n/2 (2a + (n-1) d)$, and equivalently $S_n = n/2 (a + l)$, where l is the last term added.

The second form says what is going on. $(a + l)/2$ is the average of the first and last terms, and $S_n$ is n copies of it.

Why the average works: the terms of an A.P. are evenly spaced. Pair the first with the last, the second with the second-last, and so on down. Every pair has the same total, because whatever the second term gained over the first, the second-last lost against the last.

So the whole sum behaves as though every term were that average value, and there are n of them.

The first form is the same statement with l written out as $a + (n-1) d$. Use whichever one the question hands you: if the last term is given, the second form is fewer steps.

n copies of the average of the ends. Remember that sentence and you can rebuild both forms.

The six-term sum written twice, once reversed, with every column totalling 26.
Worked example

Sum the first 10 terms of 3, 7, 11, 15, ...

  1. $a = 3$, $d = 4$, $l = a_{10} = 39$
    The pieces are already known from the previous worked example.
  2. $S_{10} = 10/2 (3 + 39) = 5 \times 42 = 210$
    The last term is in hand, so use the shorter form $S_n = n/2 (a + l)$.
  3. $S_{10} = 10/2 (2 \times 3 + 9 \times 4) = 5 \times (6 + 36) = 5 \times 42 = 210$
    Check it against the other form, which needs no l — the same.
  4. $(3+39)/2 = 21$; $5 \times 42 = 210$
    The average of the ends, ten copies of it — the sentence from the last section, arriving as arithmetic.
  5. $10 \times 3 = 30$ to $10 \times 39 = 390$
    A rough check catches gross errors — 210 sits comfortably between.
The ten terms as bars, with the dashed line at the height they would all share if the staircase were flattened.

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The arithmetic mean

●KEY-TERM

The ARITHMETIC MEAN of two numbers a and b is $(a+b)/2$. That is the ordinary average, and it also has a second description that this chapter needs.

Put $(a+b)/2$ between a and b, and the three numbers $a$, $(a+b)/2$, $b$ form an A.P.

Check that they do. The step from a to the mean is $(a+b)/2 - a$, which is $(b-a)/2$. The step from the mean to b is $b - (a+b)/2$, which is also $(b-a)/2$. Equal steps, so it is an A.P. with common difference $(b-a)/2$.

So the arithmetic mean is the number sitting an equal ADDITIVE distance from both. Not equally far in some vague sense — the same amount added, then the same amount added again.

That is the description to carry, because the geometric mean gets exactly the same description with one word changed.

The A.M. is the single term that turns two numbers into a three-term A.P.

The same interval cut twice: only the cut at the mean leaves two pieces of equal size.
Worked example

Insert the arithmetic mean between 4 and 12

  1. $(4+12)/2 = 16/2 = 8$
    Apply the definition — the claim is that 4, 8, 12 is an A.P.
  2. $8 - 4 = 4$, $12 - 8 = 4$
    Check it — the same common difference both times, so the three terms do form an A.P.
  3. $(12-4)/2 = 4$
    The common difference is half the gap between the two original numbers, exactly as the derivation predicted — and 8 lies genuinely between 4 and 12, since it is their average.
The inserted mean with both gaps drawn: 4 to 8 and 8 to 12 are the same arc.

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Geometric progressions

▲CONCEPT

A GEOMETRIC PROGRESSION, or G.P., is the multiply-instead-of-add case.

Each term after the first is the previous one multiplied by a fixed, nonzero COMMON RATIO, written r. From a first term a the sequence runs $a$, $a r$, $a r^2$, $a r^3$, and so on.

The nth term is $a_n = a r^{n-1}$.

The $(n-1)$ does the same job it did in the A.P. Reaching the nth term takes $n-1$ multiplications by r, not n, because the first term has been multiplied by nothing.

Check at the start. $n = 1$ gives $a_1 = a r^0 = a$, the first term. Correct.

Compare the two definitions directly. In an A.P. the position appears as a MULTIPLIER of d; in a G.P. it appears as an EXPONENT of r. That is the whole structural difference, and it is why a G.P. outruns an A.P. so quickly.

The ratio has to be nonzero, since an r of 0 would collapse every term after the first to zero. It may be negative, in which case the terms alternate in sign, and it may be a fraction, in which case they shrink.

Divide consecutive terms to find r. If the quotient is constant, it is a G.P.

Five G.P. terms with the multiplication arcs above them and the exponent-on-r rail below.
Worked example

Find the 5th term of 2, 6, 18, 54, ...

  1. $a = 2$; $6/2 = 3$, $18/6 = 3$, $54/18 = 3$
    Find the two constants, dividing consecutive terms to confirm r is constant — one pair can agree by accident, so check at least two.
  2. $a_5 = 2 \times 3^{5-1} = 2 \times 3^4 = 2 \times 81 = 162$
    Substitute into $a_n = a r^{n-1}$ with $n = 5$ — one power to evaluate, where walking the list is four separate multiplications.
  3. exponent 4, not 5
    Four multiplications by 3 take you from the first term to the fifth. Using 5 would give 486, the 6th term.
  4. 2, 6, 18, 54, 162
    Check by extending the list — the fifth entry is 162.
The five terms drawn to one linear scale, so the answer can be checked against the size it should be.

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Adding up a G.P.

▲CONCEPT

The sum of the first n terms of a G.P. has a formula too, with one condition attached.

For $r \neq 1$: $S_n = a(r^n - 1)/(r - 1)$. The equivalent form $a(1 - r^n)/(1 - r)$ is the same expression with top and bottom both negated, and it is tidier when r is less than 1.

The condition is not decoration. If $r = 1$ then $r - 1 = 0$, and the formula divides by zero. There is no value to salvage there; the derivation itself breaks at that point.

But the $r = 1$ case is easy on its own. A common ratio of 1 means every term equals a — the sequence is $a, a, a, \ldots$ — so the sum of n of them is $S_n = n a$. No formula needed.

So the working procedure has two steps. Check whether r is 1. If it is, the answer is $n a$. If it is not, use the general formula.

That check costs a second and prevents an undefined expression, which is a worse outcome than a wrong number. A wrong number can at least be spotted.

*Look at r before writing the formula down. $r = 1$ is a separate case, not an awkward instance of the same one.*

The five terms of a doubling G.P. laid end to end, against one bar of the term that comes next.
Worked example

Sum the first 5 terms of 2, 6, 18, 54, ...

  1. $a = 2$, $r = 3$
    The constants are known from the earlier example.
  2. $r = 3$, not 1
    Run the check first — the general formula applies.
  3. $S_5 = 2(3^5 - 1)/(3 - 1) = 2(243 - 1)/2 = 2 \times 242/2 = 242$
    Substitute into $S_n = a(r^n - 1)/(r - 1)$ with $n = 5$.
  4. $2 + 6 + 18 + 54 + 162 = 242$
    Check by adding the terms directly — the formula agrees.
  5. 162 of 242
    The last term dominates — in a G.P. with $r > 1$ the total is always close to the last term, a useful rough check.
The terms above, the running total below: the last chip and the formula both give 242.

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The geometric mean

●KEY-TERM

The GEOMETRIC MEAN of two POSITIVE numbers a and b is $\sqrt{a b}$.

It has the same description as the arithmetic mean with one word changed. Put $\sqrt{a b}$ between a and b, and the three numbers $a$, $\sqrt{a b}$, $b$ form a G.P.

Check the ratios. From a to the mean the ratio is $\sqrt{a b}/a$. From the mean to b it is $b/\sqrt{a b}$. Multiply top and bottom of that second one by $\sqrt{a b}$ and it becomes $(b \sqrt{a b})/(a b)$, which is $\sqrt{a b}/a$. The same ratio both times, so it is a G.P.

So the A.M. sits an equal ADDITIVE distance from both numbers, and the G.M. sits an equal MULTIPLICATIVE distance. That is the whole difference between them.

The positivity condition is needed. $\sqrt{a b}$ has to be a real number, and if a and b have opposite signs their product is negative.

The G.M. is the single term that turns two numbers into a three-term G.P., exactly as the A.M. does for an A.P.

The same pair, 3 and 12: equal multipliers put the mean at 6, equal gaps put it at 7.5.
Worked example

Insert the geometric mean between 4 and 9

  1. $\sqrt{4 \times 9} = \sqrt{36} = 6$
    Apply the definition — the claim is that 4, 6, 9 is a G.P.
  2. $6/4 = 3/2$, $9/6 = 3/2$
    Check the ratios, never the differences — the same both times, so the three terms do form a G.P.
  3. $6 - 4 = 2$ but $9 - 6 = 3$
    The differences are NOT equal, and that is expected — this is a G.P., not an A.P., so ratios match and differences do not.
  4. A.M. $= (4+9)/2 = 6.5$
    Contrast with the arithmetic mean of the same pair — a different number doing a different job.
A 4-by-9 rectangle beside a square of side 6, both drawn to one scale and both covering 36.

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Which mean is larger

▲CONCEPT

Both means exist for any two positive numbers, and one of them is always at least as large.

For any positive a and b: $(a+b)/2 \geq \sqrt{a b}$.

The arithmetic mean is never smaller than the geometric mean. Equality holds in exactly one case, when $a = b$. Whenever the two numbers differ at all, the A.M. is strictly the larger.

Here is why, without much algebra. The square of any real number is never negative, so $(\sqrt{a} - \sqrt{b})^2 \geq 0$. Expanding that gives $a - 2 \sqrt{a b} + b \geq 0$, which rearranges to $a + b \geq 2 \sqrt{a b}$, and halving both sides gives the result.

That derivation also explains the equality case. $(\sqrt{a} - \sqrt{b})^2$ is zero only when $\sqrt{a} = \sqrt{b}$, which for positive numbers means $a = b$.

The A.M. wins unless the two numbers are identical, and then it ties. There is no case where the G.M. comes out larger.

Two semicircles on the same diameter: one with the join off-centre, one with it at the centre.
Worked example

Compare the A.M. and G.M. of three pairs

  1. A.M. $= (4+9)/2 = 6.5$; G.M. $= \sqrt{4 \times 9} = 6$
    First take 4 and 9, which are different — the arithmetic mean is strictly the larger, as it must be whenever the two numbers differ.
  2. A.M. $= (5+5)/2 = 5$; G.M. $= \sqrt{5 \times 5} = 5$
    Now take 5 and 5, which are equal — matching the equality case exactly.
  3. A.M. $= 50.5$; G.M. $= 10$, for 1 and 100
    Push the numbers further apart and the gap widens; bring them together and it shrinks to nothing — a worked case where the G.M. came out larger would mean an arithmetic error, not a counterexample.
Both means between a and b, on one scale: the gap between them closes exactly when a equals b.

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Three sums worth knowing

▲CONCEPT

Three sums come up often enough to be worth knowing as closed forms, and none of them is an A.P. or G.P. result in the usual sense.

The sum of the first n natural numbers:

$\sum_{k=1}^n k = n(n+1)/2.$

The sum of their squares:

$\sum_{k=1}^n k^2 = n(n+1)(2n+1)/6.$

The sum of their cubes:

$\sum_{k=1}^n k^3 = (n(n+1)/2)^2.$

The first is an A.P. sum in disguise. The natural numbers have common difference 1, and the formula is the A.P. sum formula with $a = 1$, $d = 1$ and $l = n$.

The other two are not. Squares run 1, 4, 9, 16: the differences are not constant and neither are the ratios, so it is neither progression, and the formula has to be established separately.

The third carries a fact worth noticing. The sum of the first n cubes is exactly the SQUARE of the sum of the first n natural numbers. There is no A.P. or G.P. counterpart to that; it is a property of cubes.

Only the first of the three is an A.P. sum. The other two are their own results and have to be recalled, not derived from a progression.

The step-to-step gaps of the naturals against those of the squares, drawn on one scale.
A square of side 10 cut into four nested bands; band k is k wide and covers exactly k cubed.
A staircase of 1 to 5 squares and a second copy of it turned over, interlocking into a plain 5-by-6 rectangle.
Worked example

Evaluate the three special sums at n = 5

  1. $\sum_{k=1}^5 k = (5 \times 6)/2 = 30/2 = 15$
    The natural numbers first.
  2. $1+2+3+4+5 = 15$
    Check directly — correct.
  3. $\sum_{k=1}^5 k^2 = (5 \times 6 \times 11)/6 = 330/6 = 55$
    The squares — the 11 comes from $2n+1$ with $n = 5$.
  4. $1 + 4 + 9 + 16 + 25 = 55$
    Check directly — correct.
  5. $\sum_{k=1}^5 k^3 = ((5 \times 6)/2)^2 = 15^2 = 225$
    The cubes.
  6. $1 + 8 + 27 + 64 + 125 = 225$
    Check directly — correct.
  7. $15^2 = 225$
    Read the last two together — the cubes fact holds here exactly. All three formulas are this cheap to check at small n, which fixes which is which far better than re-reading them does.
The squares accumulated column by column to 55, with all three sums at n = 5 written under them.

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Two progressions, one question

■RECAP

One question runs through the chapter: does the sequence add, or does it multiply?

ADDS — an A.P., with common difference d. Its nth term is $a_n = a + (n-1) d$ and its sum is $S_n = n/2 (2a + (n-1) d)$, which is n copies of the average of the ends.

MULTIPLIES — a G.P., with common ratio r. Its nth term is $a_n = a r^{n-1}$ and its sum is $S_n = a(r^n - 1)/(r - 1)$, valid only for $r \neq 1$; at $r = 1$ the sum is $n a$.

The position offset is the same on both sides. It is $n-1$, because the first term has taken no steps and no multiplications.

THE MEANS insert one term between two numbers. The arithmetic mean $(a+b)/2$ makes a three-term A.P., and the geometric mean $\sqrt{a b}$ makes a three-term G.P. And $(a+b)/2 \geq \sqrt{a b}$ always holds, with equality only when the two numbers were already equal.

THE THREE SPECIAL SUMS stand outside all of it. The formulas for the first n natural numbers, their squares and their cubes are closed forms for their own sequences, not instances of either progression.

Identify which progression you have before reaching for any formula. Half the formulas in this chapter are wrong for the other case.

An A.P. and a G.P. started from the same 3, drawn term by term against one axis.

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Two traps this chapter sets

✕MISCONCEPTION

THE TRAP. The nth term of an A.P. is $a + n d$. Multiply the common difference by n, since the nth term is what you want.

THE REALITY. It is $a_n = a + (n-1) d$. The multiplier is $n-1$.

Test it at the very first term, where the difference is easiest to see. The correct formula gives $a_1 = a + 0 \times d = a$, the first term itself. The trap gives $a + d$, which is the second.

The reason is that the first term is a starting point, not a step. It has had no d added to it at all. The second has had one, the third two, and the nth has had $n-1$.

What makes this expensive is the shape of the wrong answer. $a + n d$ is not nonsense. It is the $(n+1)$th term — a perfectly real term of the same sequence, just one further along than the one asked for.

It is the same off-by-one as the binomial theorem’s general term, arriving from the other direction. There the formula gave $T_{r+1}$ and the position had to be reduced; here the position is right and the step count has to be.

*Substitute $n = 1$ into any general-term formula you write. It must return the first term.*

The same six terms twice: only the ringed column moves, and the trap moves it one place right.
✕MISCONCEPTION

THE TRAP. $S_n = a(r^n - 1)/(r - 1)$ is the G.P. sum formula, so use it on every G.P.

THE REALITY. It fails on one of them. If $r = 1$, the denominator $r - 1$ is 0 and the expression is undefined.

That is not a formula returning a slightly wrong number. It returns nothing at all.

Look at what $r = 1$ actually is. The sequence runs $a, a, a, \ldots$ — every term the same, since multiplying by 1 changes nothing. It is a legitimate G.P. under the definition, which asks only that r be fixed and nonzero.

And its sum needs no machinery. n copies of a is $S_n = n a$. For 5, 5, 5, 5 the sum is $4 \times 5 = 20$, and no formula was required to see it.

So the working rule has two branches and the check comes first. If $r = 1$, write $n a$. Otherwise use the general formula.

The habit generalises past this chapter. Whenever a formula carries a stated condition, that condition marks a case which has to be handled somewhere else, not a footnote to be skimmed.

Read r before writing the formula. A denominator that can be zero is a case, not a warning.

Five equal terms with the two formulas that could be reached for underneath: only one of them returns a number.

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Practice set

Exercise 8.1 — Arithmetic progressions
  1. practice Find the 15th term of the A.P. $5, 8, 11, 14, \ldots$
  2. practice Find the sum of the first 12 terms of the A.P. $2, 7, 12, \ldots$
  3. practice Insert the arithmetic mean between 7 and 21.
  4. practice In the A.P. $4, 7, 10, \ldots$, which term equals 58?
    1. 17th
    2. 18th
    3. 19th
    4. 20th
Answers
  1. 47
  2. 354
  3. 14
  4. 19th
Exercise 8.2 — Geometric progressions
  1. practice Find the 6th term of the G.P. $3, 6, 12, 24, \ldots$
  2. practice Find the sum of the first 6 terms of the G.P. $1, 3, 9, 27, \ldots$
  3. practice Find the sum of the first 8 terms of the G.P. $5, 5, 5, 5, \ldots$
Answers
  1. 96
  2. 364
  3. 40
Miscellaneous
  1. practice Insert the geometric mean between 5 and 45.
  2. practice Find the A.M. and G.M. of 16 and 4, and state which is larger.
  3. practice Find the sum of the squares of the first 6 natural numbers.
  4. practice Find the sum of the cubes of the first 4 natural numbers.
  5. practice A man saves Rs 100 in the first month and increases his savings by Rs 50 each month after that. Find his total savings over the first 10 months.
  6. practice Which of these sequences is a G.P.?
    1. $2, 4, 6, 8$
    2. $2, 4, 8, 16$
    3. $2, 4, 7, 11$
    4. $2, 4, 8, 12$
Answers
  1. 15
  2. A.M. = 10, G.M. = 8; A.M. $>$ G.M.
  3. 91
  4. 100
  5. Rs 3250
  6. $2, 4, 8, 16$
All four options open 2, 4, so any G.P. among them has ratio 2 — the dashes mark where each later term would have to land. Option D holds the ratio for three terms and drops it on the fourth, which is exactly what makes it tempting. Check every gap before answering, not the first one.

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Chapter-end problems

Chapter-end problems — graded set
  1. board-easy A sequence is defined by $a_n = n^2 - n + 1$. Find $a_5$.
  2. board-easy A sequence has $a_n = 2n+1$. Find $S_3$, the sum of its first 3 terms.
  3. board-easy An A.P. has $a=3$ and $d=5$. Find $a_{12}$.
  4. board-easy Find the 15th term of the A.P. $4, 9, 14, 19, \ldots$.
  5. board-easy Find $S_{10}$ for the A.P. with $a=2$ and $d=3$.
  6. board-easy An A.P. has first term 5 and last (10th) term 50. Find $S_{10}$.
  7. board-easy Which of these is the arithmetic mean of 18 and 32?
    1. $25$
    2. $24$
    3. $26$
    4. $23$
  8. board-easy Insert the arithmetic mean between $-6$ and $14$.
  9. board-easy A G.P. has $a=5$ and $r=2$. Find $a_6$.
  10. board-easy Find the 5th term of the G.P. $1, 3, 9, 27, \ldots$.
  11. board-easy Find $S_5$ for the G.P. with $a=3$ and $r=2$.
  12. board-easy A G.P. has $a=9$ and every term equal ($r=1$). Find $S_7$.
  13. board-easy Which of these is the geometric mean of 5 and 45?
    1. $15$
    2. $20$
    3. $25$
    4. $10$
  14. board-easy Insert the geometric mean between 8 and 18.
  15. board-easy Find the A.M. and the G.M. of 9 and 16.
  16. board-easy Find the sum of the first 6 natural numbers.
  17. board-easy Find the sum of the squares of the first 5 natural numbers.
  18. board-standard Which term of the A.P. $7, 10, 13, \ldots$ equals 100?
    1. 30th
    2. 31st
    3. 32nd
    4. 33rd
  19. board-standard An A.P. has $a_3 = 13$ and $a_7 = 29$. Find $a$ and $d$.
  20. board-standard Three numbers in A.P. have sum 24 and product 440. Find them.
  21. board-standard An A.P. has $a=2$ and $d=2$. Its sum is 110. Find $n$.
  22. board-standard Insert 3 arithmetic means between 4 and 24.
  23. board-standard A G.P. has $a_2 = 6$ and $a_5 = 48$. Find $a$ and $r$.
  24. board-standard Three numbers in G.P. have sum 21 and product 216. Find them (take $r>1$).
  25. board-standard A G.P. has $a=3$ and $r=2$. Its sum is 765. Find $n$.
  26. board-standard The sum of the first n terms of a series is $S_n = 3n^2 + 2n$. Find $a_n$.
  27. board-standard A sequence has $a_n = 5n - 2$. Confirm it is an A.P. and find $a_{10}$.
  28. board-standard A stadium’s front row seats 20; each row behind adds 4 more seats. Find the seats in row 15 and the total seats in the first 15 rows.
  29. board-standard A bacterial culture starts at 500 and doubles every hour. Find the count after 6 hours.
  30. board-standard A savings plan deposits 1000 in year 1 and 500 more each following year (an A.P.). Find the total saved over 10 years.
  31. board-standard Counting ancestors: 2 parents, 4 grandparents, 8 great-grandparents, doubling each generation (a G.P. with $a=2$, $r=2$). Find the total across 8 generations.
  32. board-standard The 20th term of the A.P. $-2, 1, 4, 7, \ldots$ is
    1. $52$
    2. $55$
    3. $58$
    4. $60$
  33. board-standard The sum of the first 8 terms of $5, 5, 5, \ldots$ is
    1. $40$
    2. $5$
    3. $13$
    4. $45$
  34. board-standard For $a=25$ and $b=4$, which is true?
    1. A.M. > G.M.
    2. A.M. < G.M.
    3. A.M. = G.M.
    4. cannot compare
  35. board-standard Find the sum of $k^2+k$ for $k=1$ to $6$, using the closed-form formulas.
  36. board-standard For $n=6$, find the sum of the cubes of the first 6 natural numbers, and confirm it equals the square of their sum.
  37. board-standard Is $3, 6, 12, 24, \ldots$ an A.P. or a G.P.? Find its 7th term.
  38. board-standard Find the sum of all natural numbers from 1 to 50 that are divisible by 3.
  39. JEE An A.P. has $S_6 = 72$ and $S_{10} = 200$. Find $a$ and $d$.
  40. JEE Three numbers in G.P. have product 1000. Adding 6 to the second and 7 to the third gives three numbers in A.P. Find the original numbers (take $r>1$).
  41. JEE A bacterial culture starts at 100 and doubles every hour. Find the smallest whole number of hours after which the count exceeds 100000.
  42. JEE Among all rectangles with perimeter 40, find the maximum possible area.
  43. JEE Find the sum of the squares of the first 10 natural numbers, minus the sum of the first 10 natural numbers.
  44. JEE Three numbers in A.P. sum to 21. Adding 2, 2, 14 to them respectively gives three numbers in G.P. Find the original numbers.
  45. JEE A G.P. has $a_1 = 8$ and $a_2 = 8$. Find $S_{12}$.
  46. JEE An investment of 8000 grows by a factor of 1.1 each year (compound growth, a G.P.). Find its value after 3 years.
  47. JEE The A.P. $3, 7, 11, \ldots, 407$ — find the number of terms and their sum.
  48. JEE For $0<x<9$, find $x$ so that the A.M. of 9 and $x$ exceeds their G.M. by exactly 2.
  49. JEE Find $1+3+5+\ldots+19$ (the first 10 odd numbers) and compare it to $10^2$.
  50. JEE An A.P. has $a_7 = 1/9$ and $a_9 = 1/7$. Find $a_{63}$.
    1. $1$
    2. $0$
    3. $1/63$
    4. $1/2$
  51. JEE A G.P. has $a_3 = 24$ and $a_6 = 192$. Find $a$, $r$, and $S_5$.
Answers
  1. $21$
  2. $15$
  3. $58$
  4. $74$
  5. $155$
  6. $275$
  7. $25$
  8. $4$
  9. $160$
  10. $81$
  11. $93$
  12. $63$
  13. $15$
  14. $12$
  15. A.M. $=25/2$, G.M. $=12$
  16. $21$
  17. $55$
  18. 32nd
  19. $a=5, d=4$
  20. $5, 8, 11$
  21. $n=10$
  22. $9, 14, 19$
  23. $a=3, r=2$
  24. $3, 6, 12$
  25. $n=8$
  26. $a_n = 6n-1$
  27. $d=5, a_{10}=48$
  28. $a_{15}=76, S_{15}=720$
  29. $32000$
  30. $32500$
  31. $510$
  32. $55$
  33. $40$
  34. A.M. > G.M.
  35. $112$
  36. $441$
  37. G.P. — $a_7=192$
  38. $408$
  39. $a=2, d=4$
  40. $5, 10, 20$
  41. $n=10$
  42. $100$
  43. $330$
  44. $1, 7, 13$
  45. $96$
  46. $10648$
  47. $n=102, S=20910$
  48. $x=1$
  49. $100 = 10^2$
  50. $1$
  51. $a=6, r=2, S_5=186$
The 12th term of the A.P. 5, 9, 13, ..., with the two off-by-one answers either side of it.
The 6th term of the G.P. 4, 12, 36, ..., with the two off-by-one answers either side of it.
JEE-tier practice — from the item bank
  1. An A.P. has first term 5 and common difference 4. Find its 12th term.
    1. 49
    2. 53
    3. 45
    4. 48
    Check your answer
    1. ✓ 49 — (A) $a_12$ = a + (12-1)d = 5 + 11(4) = 49 — eleven steps of d from the first term.
    2. 53 — The nth term needs (n-1) steps of d from a, not n steps. 5 + 12(4) = 53 lands one term too far, on the 13th term.
    3. 45 — This is 5 + 10(4) = 45, only 10 steps — one step short of the 11 needed to reach the 12th term.
    4. 48 — 12(4) = 48 drops the first term a entirely. The first term itself is part of every term.
  2. An A.P. has first term -3 and common difference 7. Find its 9th term.
    1. 60
    2. 53
    3. 46
    4. 63
    Check your answer
    1. 60 — -3 + 9(7) = 60 uses 9 steps of d, one too many for the 9th term — that lands on the 10th term.
    2. ✓ 53 — (B) $a_9$ = -3 + (9-1)(7) = -3 + 56 = 53 — eight steps of d from the first term.
    3. 46 — -3 + 7(7) = 46 uses only 7 steps, one short of the 8 needed.
    4. 63 — 9(7) = 63 drops the first term a entirely.
  3. Find the sum of the first 15 terms of the A.P. 4, 10, 16, ....
    1. 735
    2. 1380
    3. 690
    4. 644
    Check your answer
    1. 735 — 15/2(2(4)+15(6)) = 735 uses 15 steps of d inside the bracket instead of 14 — one term too far.
    2. 1380 — 15(2(4)+14(6)) = 1380 is exactly double the correct sum — the /2 in the formula was dropped.
    3. ✓ 690 — (C) $S_15$ = 15/2(2(4)+14(6)) = 15/2(8+84) = 15/2(92) = 690.
    4. 644 — 14/2(2(4)+14(6)) = 644 uses the leading factor for 14 terms, not 15.
One arithmetic progression, its nth term and its running total plotted across fifteen terms.
JEE-tier practice — continued
  1. Find the sum of the first 9 terms of the A.P. 6, 12, 18, ....
    1. 297
    2. 540
    3. 486
    4. 270
    Check your answer
    1. 297 — 9/2(2(6)+9(6)) = 297 uses 9 steps of d inside the bracket instead of 8.
    2. 540 — 9(2(6)+8(6)) = 540 is exactly double the correct sum — the /2 was dropped.
    3. 486 — 9/2(2(6)+8(12)) = 486 doubles d to 12 inside the bracket by mistake.
    4. ✓ 270 — (D) $S_9$ = 9/2(2(6)+8(6)) = 9/2(12+48) = 9/2(60) = 270 — the sum of the first 9 multiples of 6.
  2. A G.P. has first term 4 and common ratio 3. Find its 6th term.
    1. 972
    2. 2916
    3. 324
    4. 243
    Check your answer
    1. ✓ 972 — (A) $a_6$ = a $r^{6-1}$ = $4(3^5)$ = 4(243) = 972 — five multiplications by r from the first term.
    2. 2916 — $4(3^6)$ = 2916 multiplies by r six times instead of five — that lands on the 7th term.
    3. 324 — $4(3^4)$ = 324 multiplies by r only four times, one short of the five needed.
    4. 243 — $3^5$ = 243 drops the first term a entirely.
  3. A G.P. has first term 81 and common ratio 1/3. Find its 5th term.
    1. 1/3
    2. 1
    3. 3
    4. 1/81
    Check your answer
    1. 1/3 — $81(1/3)^5$ = 1/3 multiplies by r five times instead of four — one term too far.
    2. ✓ 1 — (B) $a_5$ = $81(1/3)^{5-1}$ = $81(1/3)^4$ = 81/81 = 1 — four multiplications by r.
    3. 3 — $81(1/3)^3$ = 3 multiplies by r only three times, one short of the four needed.
    4. 1/81 — $(1/3)^4$ = 1/81 drops the first term a entirely.
The same two curves for a geometric progression, where the total stays close to the last term.
JEE-tier practice — continued
  1. Find the sum of the first 4 terms of the G.P. 2, 10, 50, 250, ....
    1. 312
    2. 1562
    3. 156
    4. 312.5
    Check your answer
    1. ✓ 312 — (A) $S_4$ = $a(r^4-1)/(r-1)$ = 2(625-1)/4 = 2(624)/4 = 312.
    2. 1562 — $2(5^5-1)/(5-1)$ = 1562 uses $r^5$ inside the formula instead of $r^4$ — one power too far.
    3. 156 — $(5^4-1)/(5-1)$ = 156 is the bracket alone; the leading factor a=2 was never applied.
    4. 312.5 — $2(5^4)/(5-1)$ = 312.5 drops the -1 from the numerator, so it never subtracts off the missing term.
  2. A G.P. has first term 12 and every term equal (common ratio 1). Find the sum of the first 9 terms.
    1. 0
    2. 108
    3. 12
    4. 120
    Check your answer
    1. 0 — The general formula $a(r^n-1)/(r-1)$ divides by zero at r=1 — it cannot be used here at all, not even to get 0.
    2. ✓ 108 — (B) At r=1 every term already equals a, so $S_9$ = 9(12) = 108 — no formula, just count and multiply.
    3. 12 — 12 reports the first term alone; every one of the 9 equal terms must be counted.
    4. 120 — 12(10) = 120 counts 10 terms instead of the 9 asked for.
  3. A G.P. has first term -7 and every term equal (common ratio 1). Find the sum of the first 6 terms.
    1. -42
    2. 0
    3. -7
    4. -49
    Check your answer
    1. ✓ -42 — (A) At r=1 every term already equals a, so $S_6$ = 6(-7) = -42.
    2. 0 — The general formula divides by zero at r=1 — it cannot produce any answer here, let alone 0.
    3. -7 — -7 reports the first term alone; all 6 equal terms must be counted.
    4. -49 — -7(7) = -49 counts 7 terms instead of the 6 asked for.
  4. Find the arithmetic mean of 11 and 39.
    1. 50
    2. 25
    3. 28
    4. 14
    Check your answer
    1. 50 — 11+39 = 50 is the sum alone — the mean still needs dividing by 2.
    2. ✓ 25 — (B) (11+39)/2 = 50/2 = 25.
    3. 28 — 28 does not match (11+39)/2 under any standard slip; recompute the sum first.
    4. 14 — (39-11)/2 = 14 averages the DIFFERENCE, not the two numbers themselves.
  5. Three numbers in A.P. have an average of 20. Find their sum.
    1. 60
    2. 20
    3. 40
    4. 20/3
    Check your answer
    1. ✓ 60 — (A) The middle term of an A.P. is the average of the outer two, and equals the mean of all three, so sum = 3(20) = 60.
    2. 20 — 20 is the average of the three numbers, not their total — the question asks for the sum.
    3. 40 — 2(20) = 40 doubles the average; three numbers need tripling, not doubling.
    4. 20/3 — 20/3 divides instead of multiplying — the sum of three numbers must be larger than their average, not smaller.
  6. Find the geometric mean of 6 and 54.
    1. 30
    2. 324
    3. 9
    4. 18
    Check your answer
    1. 30 — (6+54)/2 = 30 is the ARITHMETIC mean — the geometric mean needs the product’s square root instead.
    2. 324 — 6(54) = 324 is the product alone; the geometric mean still needs the square root of it.
    3. 9 — 9 does not match $\sqrt{6(54)}$; recompute the product before taking the root.
    4. ✓ 18 — (D) $\sqrt{6(54)}$ = $\sqrt{324}$ = 18.
  7. Find the geometric mean of 16 and 25.
    1. 20
    2. 20.5
    3. 400
    4. 1
    Check your answer
    1. ✓ 20 — (A) $\sqrt{16(25)}$ = $\sqrt{400}$ = 20.
    2. 20.5 — (16+25)/2 = 20.5 is the ARITHMETIC mean — the geometric mean needs the product’s square root instead.
    3. 400 — 16(25) = 400 is the product alone; the geometric mean still needs the square root of it.
    4. 1 — $\sqrt{25}-\sqrt{16}$ = 1 subtracts the two roots — the geometric mean multiplies under one shared root instead.
  8. By how much does the arithmetic mean of 49 and 9 exceed their geometric mean?
    1. 29
    2. 8
    3. 21
    4. 50
    Check your answer
    1. 29 — 29 is the A.M. itself, (49+9)/2 — the question asks how much it EXCEEDS the G.M., not its own value.
    2. ✓ 8 — (B) A.M. = (49+9)/2 = 29, G.M. = $\sqrt{49(9)}$ = 21, so the A.M. exceeds the G.M. by 29-21 = 8.
    3. 21 — 21 is the G.M. itself, $\sqrt{49(9)}$ — the question asks for the gap above it, not its own value.
    4. 50 — 29+21 = 50 adds the two means; the question asks for their difference instead.
  9. For $a=25$, for which value of $b$ does the arithmetic mean of $a$ and $b$ equal their geometric mean?
    1. 0
    2. 5
    3. 25
    4. 50
    Check your answer
    1. 0 — b=0 fails immediately — the geometric mean needs both numbers POSITIVE, and the A.M. would exceed the G.M. anyway (12.5 vs 0).
    2. 5 — 5 is $\sqrt{25}$, not the value of b — equality in A.M. >= G.M. holds only when the two numbers are themselves equal.
    3. ✓ 25 — (C) A.M. >= G.M. holds with equality exactly when a = b, so b = 25 (matching A.M. = G.M. = 25).
    4. 50 — 50 is 2a, not the equality condition — check: A.M.=(25+50)/2=37.5, $G \cdot M \cdot =\sqrt{1250}$ is about 35.4, not equal.
  10. Find the sum of the first 12 natural numbers.
    1. 144
    2. 78
    3. 66
    4. 156
    Check your answer
    1. 144 — $12^2$ = 144 squares n directly — the correct closed form is n(n+1)/2, not $n^2.$
    2. ✓ 78 — (B) sum = n(n+1)/2 = 12(13)/2 = 156/2 = 78.
    3. 66 — 11(12)/2 = 66 uses n=11 inside the formula, one short of the 12 needed.
    4. 156 — 12(13) = 156 is exactly double the correct sum — the /2 was dropped.
  11. Find the sum of the squares of the first 8 natural numbers.
    1. 204
    2. 1296
    3. 180
    4. 1224
    Check your answer
    1. ✓ 204 — (A) sum of squares = n(n+1)(2n+1)/6 = 8(9)(17)/6 = 1224/6 = 204.
    2. 1296 — 1296 = $(8(9)/2)^2$ is the SUM-OF-CUBES identity (the square of the sum of naturals), not the sum of squares.
    3. 180 — 8(9)(15)/6 = 180 uses (2n-1)=15 instead of (2n+1)=17 inside the formula.
    4. 1224 — 8(9)(17) = 1224 is the numerator alone — the /6 was dropped.
  12. Find the sum of the cubes of the first 3 natural numbers.
    1. 6
    2. 14
    3. 36
    4. 27
    Check your answer
    1. 6 — 3(4)/2 = 6 is the sum of the first 3 natural numbers, not their cubes.
    2. 14 — 3(4)(7)/6 = 14 is the sum of their SQUARES, not their cubes.
    3. ✓ 36 — (C) sum of cubes = $(n(n+1)/2)^2$ = $(3(4)/2)^2$ = $6^2$ = 36 — the square of the sum of naturals.
    4. 27 — $3^3$ = 27 cubes only the last term; all three terms’ cubes must be added together.
  13. The sequence 7, 7, 7, 7, ... is both an A.P. (common difference 0) and a G.P. (common ratio 1). Find the sum of its first 6 terms.
    1. 7
    2. 42
    3. 0
    4. 49
    Check your answer
    1. 7 — 7 reports one term alone; all 6 equal terms in the sum must be counted.
    2. ✓ 42 — (B) Every term already equals 7, so $S_6$ = 6(7) = 42 — read as an A.P. with d=0 or a G.P. with r=1, both give the same answer.
    3. 0 — The general G.P. sum formula divides by zero at r=1 — it produces no valid answer here, let alone 0.
    4. 49 — 7(7) = 49 counts 7 terms instead of the 6 asked for.
  14. A quantity follows the G.P. 250, 1000, 4000, .... Find the first term of the sequence that exceeds 60000.
    1. 16000
    2. 64000
    3. 256000
    4. 60000
    Check your answer
    1. 16000 — 16000 is the 4th term — it is below 60000, so it does not yet exceed the threshold.
    2. ✓ 64000 — (B) The terms are 250, 1000, 4000, 16000, 64000, .... The 4th term (16000) is still below 60000; the 5th term, 64000, is the first to exceed it.
    3. 256000 — 256000 is the 6th term — the sequence had already exceeded 60000 one term earlier, at the 5th term.
    4. 60000 — 60000 is the threshold itself, not a term of the sequence — no term equals it exactly.
The two given terms are three columns apart, so the ratio comes from a cube root, not a fifth root.
Four rectangles on the same 40 of fence: the sides always average 10, so the area never beats 10 squared.
Each step multiplies by 4, so the terms never approach 60000 — the 4th sits 44000 below it, the 5th lands 4000 above. When an item asks for the first term past a threshold, find the last one below and take the next: nothing sits in between.

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