"Sequences and Series" — Practice

Question 1

The sum of three numbers in arithmetic progression is 27 and their product is 648. Find the numbers.

Accepted Answer

Choose the terms symmetrically as a minus d, a, and a plus d. The sum is 3a because the d terms cancel, so 3a equals 27 and a equals 9. The product is a times the quantity a squared minus d squared, using the difference of squares on the outer pair. That gives 9 times (81 minus d squared) equal to 648, so 81 minus d squared equals 72 and d squared equals 9, giving d equal to plus or minus 3. Either sign yields the same set in one order or the other, namely 6, 9 and 12. Checking, they sum to 27 and multiply to 648. Writing the terms as a, a plus d, a plus 2d would have carried both unknowns into the product equation.

Question 2

Insert five arithmetic means between 8 and 26, and verify their sum against the general result.

Accepted Answer

Inserting five means gives seven terms in total, with 8 first and 26 last. Seven terms means six steps, so the common difference is 26 minus 8 all over 6, which is 3. The inserted means are 11, 14, 17, 20 and 23, and their sum is 85. The general result predicts n times the single arithmetic mean, which is 5 times the average of 8 and 26, that is 5 times 17, giving 85. The two agree, and the reason is visible in the list: the means are symmetric about 17, so their average is 17 and their sum is five times that.

Question 3

How many terms of the arithmetic progression 3, 7, 11, ... must be taken for the sum to be 210?

Accepted Answer

Here the first term is 3 and the common difference is 4. The sum of n terms is n over 2 times the quantity 2 times 3 plus (n minus 1) times 4, which simplifies to n over 2 times (4n plus 2), that is n times (2n plus 1). Setting this equal to 210 gives 2n squared plus n minus 210 equal to zero. The discriminant is 1 plus 1680, which is 1681, whose square root is 41. So n equals minus 1 plus or minus 41, all over 4, giving n equal to 10 or n equal to minus 10.5. A number of terms must be a positive whole number, so n equals 10. Checking, the tenth term is 39 and the sum of 3 through 39 is 10 times the average 21, which is 210.

Question 4

Express the recurring decimal 0.454545... as a fraction using an infinite geometric series, and state why the method is valid.

Accepted Answer

Write the decimal as 0.45 plus 0.0045 plus 0.000045 and so on. This is a geometric progression whose first term is 45 over 100 and whose common ratio is 1 over 100, since each block is shifted two more places to the right. The magnitude of the ratio is less than one, which is precisely the condition for the infinite sum to exist, so the formula applies. The sum is (45 over 100) divided by (1 minus 1 over 100), which is (45 over 100) divided by (99 over 100), giving 45 over 99, which reduces to 5 over 11. Dividing 5 by 11 returns 0.4545 and so on, confirming the answer.

Question 5

Insert three geometric means between 2 and 32, and verify their product against the general result.

Accepted Answer

Three means gives five terms in total, with 2 first and 32 last. Five terms means four multiplications, so r to the fourth equals 32 over 2, which is 16, giving r equal to 2 for the positive progression. The inserted means are 4, 8 and 16, and their product is 512. The general result predicts the nth power of the single geometric mean, which is the square root of 2 times 32, that is 8, raised to the third power, giving 512. The two agree. Compare this with inserting arithmetic means, where the means added to n times the single AM; here they multiply to the nth power, which is the same statement seen through logarithms.

Question 6

The product of three numbers in geometric progression is 216 and the sum of their products taken in pairs is 156. Find the numbers.

Accepted Answer

Take the terms as a over r, a, and a times r, so that the product collapses. The product is a cubed equal to 216, giving a equal to 6 immediately with no reference to r. The products in pairs are a squared over r, a squared times r, and a squared, whose sum is a squared times the quantity r plus one over r plus one. Substituting a squared equal to 36 gives 36 times that bracket equal to 156, so r plus one over r plus one equals 13 over 3, hence r plus one over r equals 10 over 3. Clearing fractions gives 3r squared minus 10r plus 3 equal to zero, whose roots are 3 and one third. Both give the same set in opposite orders, namely 2, 6 and 18. Checking, the product is 216 and the pairwise products are 12 plus 108 plus 36, which is 156.

Question 7

A geometric progression has first term 5 and the sum of its first n terms is 5n. What is the common ratio, and what does this reveal about a trap in the sum formula?

Accepted Answer

If the sum of n terms is 5n for every n, then each term must equal 5, since the sum increases by exactly 5 each time a term is added. A progression in which every term is the same has common ratio 1. This exposes the trap: the standard sum formula divides by r minus one, so it is undefined precisely in this case and cannot be used. The correct statement is that when r equals 1 the sum of n terms is n times the first term, which here is 5n as given. Questions often conceal r equal to 1 inside an algebraic condition, and a candidate who applies the formula without checking the denominator loses the whole question.

Question 8

Find the minimum value of x plus 9 over x for x greater than zero, and the value of x at which it occurs.

Accepted Answer

First check that AM-GM will be useful by computing the product of the two parts: x times 9 over x is 9, which is constant, so the method applies. By AM-GM on the two positive quantities, half of their sum is at least the square root of their product, which is 3. Multiplying by two gives x plus 9 over x at least 6. Equality in AM-GM holds exactly when the two quantities are equal, so x equals 9 over x, giving x squared equal to 9 and x equal to 3, taking the positive root as the domain requires. The minimum value is 6, attained at x equal to 3. Calculus gives the same answer but requires differentiating, solving and testing the second derivative.

Question 9

Two positive numbers have arithmetic mean 10 and geometric mean 8. Find the numbers, and explain why the discriminant was guaranteed to be non-negative.

Accepted Answer

The two numbers have sum equal to twice the arithmetic mean, which is 20, and product equal to the square of the geometric mean, which is 64. Those are exactly the symmetric functions that determine a quadratic, so the numbers are the roots of x squared minus 20x plus 64 equal to zero. The discriminant is 400 minus 256, which is 144, so the roots are 20 plus or minus 12, all over 2, giving 16 and 4. Checking, their arithmetic mean is 10 and their geometric mean is the square root of 64, which is 8. The discriminant is four times the quantity A squared minus G squared, and since AM-GM guarantees A is at least G for positive numbers, that quantity cannot be negative. Equivalently, if a question ever supplies a geometric mean larger than the arithmetic mean, no such pair of positive numbers exists.

Question 10

Prove the AM-GM inequality for two positive numbers, and state precisely when equality holds.

Accepted Answer

Let a and b be positive. Consider the square of the difference of their square roots, that is the square of root a minus root b. A square of a real number is never negative, so this quantity is at least zero. Expanding it gives a minus twice the square root of ab plus b, which is at least zero. Rearranging gives a plus b at least twice the square root of ab, and dividing by two gives the arithmetic mean at least the geometric mean, which is the required inequality. Equality holds exactly when the square that we started from is zero, that is when root a equals root b, which for positive numbers means a equals b. So the two means coincide precisely when the two numbers are equal, and never otherwise.

Question 11

A student applies AM-GM to x plus 1 over x squared for x greater than zero and concludes the minimum is 2. Identify the error.

Accepted Answer

The error is that the product of the two parts is not constant. Multiplying x by 1 over x squared gives 1 over x, which still depends on x. AM-GM does give that the sum is at least twice the square root of 1 over x, but that lower bound changes as x changes, so it establishes nothing about a fixed minimum. The inequality itself is still true; it is simply not useful in this form. To use AM-GM here the expression must first be regrouped so that the parts have a constant product, for example by splitting x into two halves so that the three parts multiply to a constant, or the question must be handled with calculus instead.

Question 12

Show that the terms equidistant from the two ends of a finite arithmetic progression have a constant sum, and state the corresponding property for a geometric progression.

Accepted Answer

Let the progression have n terms, first term a and common difference d. The kth term from the beginning is a plus (k minus 1)d, and the kth term from the end is the last term minus (k minus 1)d, that is a plus (n minus 1)d minus (k minus 1)d. Adding the two, the terms in (k minus 1)d cancel exactly, leaving 2a plus (n minus 1)d, which does not involve k. So the sum is the same for every k, and in particular equals the sum of the first and last terms. The corresponding GP property replaces addition by multiplication: the kth term from each end multiply to a constant, namely the product of the first and last terms. For five terms this means the first times the fifth equals the square of the middle term.

Practice — "Sequences and Series"

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