Sequences & Series

8 methods · 14 questions

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Conditions on a geometric progression routinely produce a quadratic or higher polynomial with multiple roots. Never accept a root blindly: a spurious root usually appears alongside the genuine one. Read the question's constraints and keep only the root that survives them.

Trigger: you reach a polynomial in the ratio $r$ or an unknown parameter $p$ , and the stem mentions convergence, positive terms, or a specific sign.

Instances:
(i) convergence of an infinite sum requires $∣ r ∣ < 1$ , rejecting any root $\geq 1$ or $\leq - 1$ ;
(ii) a condition that all terms are positive rejects a root that makes the first term negative or forces a negative ratio;
(iii) a term stated as 'real and positive' uniquely selects the positive root of an even power.

Linked questions (3)

NSAA 2016 Section 1, Part E (Mathematics questions) · Q77

The first term of a convergent geometric series is 8.

The fifth term is 2.

The sixth term is real and positive.

What is the sum to infinity of this series?

(The sum to infinity of a convergent geometric series is given by

\frac{a}{1 - r}

, where

a

is the first term and

r

is the common ratio.)

A $8 (1 + 2)$
B $8 (1 - 2)$
C $8 (2 + 2)$
D $8 (2 - 2)$
E $16$
F $\frac{8 5 4}{5 4 - 1}$
G $\frac{8 5 4}{5 4 + 1}$

Worked solution

Answer:C

Let

r

be the common ratio. The first term is

a = 8

and the fifth term is

a r^{4}

, so

8 r^{4} = 2, r^{4} = \frac{1}{4} .

The real solutions are

r = \pm \frac{1}{2}

. The sixth term is

8 r^{5}

, which has the same sign as

r

, so the requirement that it be positive forces

r = \frac{1}{2} .

This value satisfies

∣ r ∣ < 1

, in keeping with the series being convergent. Note that the fifth term is

a r^{4}

and not

a r^{5}

, which is why the exponent is

4

.

Hence

S_{\infty} = \frac{a}{1 - r} = \frac{8}{1 - \frac{1}{2}} = \frac{8 2}{2 - 1} .

Rationalise by multiplying numerator and denominator by

2 + 1

, noting that

(2 - 1) (2 + 1) = 1

S_{\infty} = 82 (2 + 1) = 8 (2 + 2) .

Numerically this is about

27.3

, which is consistent with the terms

8

5.66

4

2.83

2

, and so on.

The answer is C.

ENGAA 2018 Section 1, Part B (Mathematics questions) · Q39

The first three terms of a geometric progression, whose terms are all greater than zero, are

(p - 2)

(2 p + 2)

and

(5 p + 14)

What is the fifth term of the progression?

A $324$
B $486$
C $1250$
D $1458$
E $3888$

Worked solution

Answer:B

In a geometric progression the ratio of consecutive terms is the same, so

\frac{2 p + 2}{p - 2} = \frac{5 p + 14}{2 p + 2}, that is (2 p + 2)^{2} = (p - 2) (5 p + 14) .

Expanding both sides gives

4 p^{2} + 8 p + 4 = 5 p^{2} + 4 p - 28

, hence

p^{2} - 4 p - 32 = 0, (p - 8) (p + 4) = 0.

The value

p = - 4

produces the three terms

- 6

- 6

- 6

, which are not greater than zero, so it is rejected by the condition in the question. Therefore

p = 8

.

The terms are then

6

18

54

, so the first term is

6

and the common ratio is

3

. The fifth term is

6 \times 3^{4} = 6 \times 81 = 486.

The answer is B.

ENGAA 2020 Section 1, Part B (Mathematics questions) · Q31

P

and

Q

are two different geometric progressions.

The 3rd term of each geometric progression is 4.

The 5th term of each geometric progression is 2.

What is the modulus of the difference between the sums to infinity of

P

and

Q

A $0$
B $8$
C $82$
D $16$
E $162$
F $32$
G $322$

Worked solution

Answer:E

Let a progression have first term

a

and common ratio

r

. The third and fifth terms give

a r^{2} = 4, a r^{4} = 2.

Dividing the second equation by the first gives

r^{2} = \frac{1}{2}

, so

r = \frac{1}{2}

r = - \frac{1}{2}

. These are the two different progressions

P

and

Q

, and in both cases

a = \frac{4}{r ^{2}} = 8

.

Both ratios satisfy

∣ r ∣ < 1

, so both sums to infinity exist. Using

S = \frac{a}{1 - r}

and rationalising each denominator:

S_{+} = \frac{8}{1 - \frac{1}{2}} = \frac{8 2}{2 - 1} = 82 (2 + 1) = 16 + 82,

S_{-} = \frac{8}{1 + \frac{1}{2}} = \frac{8 2}{2 + 1} = 82 (2 - 1) = 16 - 82 .

Their difference is

S_{+} - S_{-} = 162

, whose modulus is

162

.

The answer is E.