NCERT Class 11 Mathematics — Sequences and Series, Exercise 8.2. All 32 questions solved.

A geometric progression has a constant ratio between consecutive terms. With first term $a$ and common ratio $r$:

$$a,\ ar,\ ar^2,\ ar^3,\ \ldots \qquad a_n = ar^{n-1}$$

$$S_n = \frac{a(r^n – 1)}{r – 1} = \frac{a(1 – r^n)}{1 – r} \quad (r \ne 1)$$

The geometric mean of $a$ and $b$ is $\sqrt{ab}$, and three numbers are in GP exactly when the middle one squared equals the product of the outer two:

$$b^2 = ac$$

Key insight. When three or four terms of a GP are unknown, choose the symmetric parametrisation: $\dfrac{a}{r},\ a,\ ar$ for three terms, because their product is simply $a^3$ and the unknowns collapse. Question 12 gives the product as $1$, so $a = 1$ immediately, and only $r$ is left to find. Picking $a, ar, ar^2$ instead leaves two unknowns entangled in both equations.

Question 1

Find the $20$th and $n$th terms of the GP $\dfrac52, \dfrac54, \dfrac58, \ldots$

Solution. Here $a = \dfrac52$ and $r = \dfrac{5/4}{5/2} = \dfrac12$.

$$a_n = ar^{n-1} = \frac52 \left(\frac12\right)^{n-1} = \frac{5}{2^n}$$

$$a_{20} = \frac{5}{2^{20}}$$

Question 2

Find the $12$th term of a GP whose $8$th term is $192$ and the common ratio is $2$.

Solution. Rather than finding $a$, step forward four terms from the $8$th — each step multiplies by $r$:

$$a_{12} = a_8 \times r^4 = 192 \times 2^4 = 192 \times 16 = 3072$$

Question 3

The $5$th, $8$th and $11$th terms of a GP are $p$, $q$ and $s$ respectively. Show that $q^2 = ps$.

Solution. Write each in terms of $a$ and $r$:

$$p = ar^4, \qquad q = ar^7, \qquad s = ar^{10}$$

Then

$$ps = (ar^4)(ar^{10}) = a^2 r^{14}, \qquad q^2 = (ar^7)^2 = a^2r^{14}$$

so $q^2 = ps$. $\blacksquare$

The reason it works: $5$, $8$, $11$ are equally spaced, so $q$ is the geometric mean of $p$ and $s$.

Question 4

The $4$th term of a GP is the square of its second term, and the first term is $-3$. Determine its $7$th term.

Solution. With $a = -3$:

$$a_2 = -3r, \qquad a_4 = -3r^3$$

The condition $a_4 = (a_2)^2$ gives

$$-3r^3 = (-3r)^2 = 9r^2$$

Dividing by $3r^2$ (non-zero, since $r = 0$ would not give a GP):

$$-r = 3 \quad\Longrightarrow\quad r = -3$$

Therefore

$$a_7 = ar^6 = -3(-3)^6 = -3 \times 729 = -2187$$

Note $(-3)^6$ is positive — an even power — so the negative sign comes entirely from $a$.

Question 5

Which term of the following sequences:

(a) $2, 2\sqrt2, 4, \ldots$ is $128$?   (b) $\sqrt3, 3, 3\sqrt3, \ldots$ is $729$?   (c) $\dfrac13, \dfrac19, \dfrac{1}{27}, \ldots$ is $\dfrac{1}{19683}$?

Solution. In each case write everything as a power of a single base and equate exponents.

(a) $a = 2$, $r = \sqrt2 = 2^{1/2}$, so $a_n = 2 \cdot 2^{(n-1)/2} = 2^{(n+1)/2}$. Setting this equal to $128 = 2^7$:

$$\frac{n+1}{2} = 7 \quad\Longrightarrow\quad n = 13$$

(b) $a = \sqrt3 = 3^{1/2}$, $r = \sqrt3$, so $a_n = 3^{n/2}$. With $729 = 3^6$:

$$\frac n2 = 6 \quad\Longrightarrow\quad n = 12$$

(c) $a = \dfrac13$, $r = \dfrac13$, so $a_n = \left(\dfrac13\right)^n$. Since $19683 = 3^9$:

$$n = 9$$

Question 6

For what values of $x$ are the numbers $-\dfrac27,\ x,\ -\dfrac72$ in GP?

Solution. Three numbers are in GP when the square of the middle equals the product of the outer two:

$$x^2 = \left(-\frac27\right)\left(-\frac72\right) = 1$$

$$x = \pm 1$$

Both values work: with $x = 1$ the ratio is $-\tfrac72$, and with $x = -1$ it is $\tfrac72$. Neither is excluded, so the answer is genuinely two-valued.

Find the sum to the indicated number of terms in each of the geometric progressions in questions 7 to 10.

Question 7

$0.15,\ 0.015,\ 0.0015,\ \ldots$ to $20$ terms

Solution. $a = 0.15$, $r = 0.1$, $n = 20$. Since $|r| < 1$, use the form with $1 – r$ on the bottom:

$$S_{20} = \frac{0.15\left(1 – (0.1)^{20}\right)}{1 – 0.1} = \frac{0.15}{0.9}\left(1 – (0.1)^{20}\right) = \frac16\left(1 – (0.1)^{20}\right)$$

Question 8

$\sqrt7,\ \sqrt{21},\ 3\sqrt7,\ \ldots$ to $n$ terms

Solution. The ratio is

$$r = \frac{\sqrt{21}}{\sqrt7} = \sqrt3$$

so

$$S_n = \frac{\sqrt7\left((\sqrt3)^n – 1\right)}{\sqrt3 – 1}$$

Rationalise the denominator by multiplying above and below by $\sqrt3 + 1$:

$$S_n = \frac{\sqrt7(\sqrt3+1)\left(3^{n/2} – 1\right)}{(\sqrt3-1)(\sqrt3+1)} = \frac{\sqrt7}{2}(\sqrt3+1)\left(3^{n/2} – 1\right)$$

Question 9

$1,\ -a,\ a^2,\ -a^3,\ \ldots$ to $n$ terms (if $a \ne -1$)

Solution. The ratio is $-a$, and the first term is $1$:

$$S_n = \frac{1 – (-a)^n}{1 – (-a)} = \frac{1 – (-a)^n}{1 + a}$$

The exclusion $a \ne -1$ is exactly what keeps the denominator from vanishing.

Question 10

$x^3,\ x^5,\ x^7,\ \ldots$ to $n$ terms (if $x \ne \pm1$)

Solution. $a = x^3$ and $r = x^2$:

$$S_n = \frac{x^3\left(1 – (x^2)^n\right)}{1 – x^2} = \frac{x^3\left(1 – x^{2n}\right)}{1 – x^2}$$

Question 11

Evaluate $\displaystyle\sum_{k=1}^{11}\left(2 + 3^k\right)$.

Solution. Split the sum into two pieces:

$$\sum_{k=1}^{11} 2 + \sum_{k=1}^{11} 3^k$$

The first is $2$ added eleven times, that is $22$. The second is a GP with $a = 3$, $r = 3$, $n = 11$:

$$\sum_{k=1}^{11} 3^k = \frac{3(3^{11} – 1)}{3 – 1} = \frac32\left(3^{11} – 1\right)$$

$$\sum_{k=1}^{11}(2 + 3^k) = 22 + \frac32\left(3^{11} – 1\right)$$

Question 12

The sum of the first three terms of a GP is $\dfrac{39}{10}$ and their product is $1$. Find the common ratio and the terms.

Solution. Take the three terms symmetrically as $\dfrac ar,\ a,\ ar$. Then the product is

$$\frac ar \cdot a \cdot ar = a^3 = 1 \quad\Longrightarrow\quad a = 1$$

and the sum condition becomes

$$\frac1r + 1 + r = \frac{39}{10} \quad\Longrightarrow\quad r + \frac1r = \frac{29}{10}$$

Multiplying by $10r$:

$$10r^2 – 29r + 10 = 0$$

$$r = \frac{29 \pm \sqrt{841 – 400}}{20} = \frac{29 \pm 21}{20} = \frac52 \ \text{ or } \ \frac25$$

Both are genuine, giving the same three numbers in opposite orders:

$$r = \frac52: \quad \frac25,\ 1,\ \frac52 \qquad\qquad r = \frac25: \quad \frac52,\ 1,\ \frac25$$

Question 13

How many terms of the GP $3, 3^2, 3^3, \ldots$ are needed to give the sum $120$?

Solution. $a = 3$, $r = 3$:

$$S_n = \frac{3(3^n – 1)}{3 – 1} = \frac32(3^n – 1) = 120$$

$$3^n – 1 = 80 \quad\Longrightarrow\quad 3^n = 81 = 3^4 \quad\Longrightarrow\quad n = 4$$

Check: $3 + 9 + 27 + 81 = 120$ ✓

Question 14

The sum of the first three terms of a GP is $16$ and the sum of the next three terms is $128$. Determine the first term, the common ratio and the sum to $n$ terms of the GP.

Solution. The next three terms are the first three multiplied by $r^3$:

$$a + ar + ar^2 = 16, \qquad ar^3 + ar^4 + ar^5 = r^3(a + ar + ar^2) = 128$$

Dividing the second by the first:

$$r^3 = \frac{128}{16} = 8 \quad\Longrightarrow\quad r = 2$$

Substituting back:

$$a(1 + 2 + 4) = 16 \quad\Longrightarrow\quad 7a = 16 \quad\Longrightarrow\quad a = \frac{16}{7}$$

$$S_n = \frac{a(r^n – 1)}{r-1} = \frac{\frac{16}{7}(2^n – 1)}{1} = \frac{16}{7}\left(2^n – 1\right)$$

Question 15

Given a GP with $a = 729$ and $7$th term $64$, determine $S_7$.

Solution.

$$a_7 = ar^6 = 729\,r^6 = 64 \quad\Longrightarrow\quad r^6 = \frac{64}{729} = \left(\frac23\right)^6$$

An even power, so both signs are possible:

$$r = \frac23 \quad\text{or}\quad r = -\frac23$$

For $r = \tfrac23$:

$$S_7 = \frac{729\left(1 – (2/3)^7\right)}{1 – 2/3} = 3 \times 729\left(1 – \frac{128}{2187}\right) = 2187 – 128 = 2059$$

For $r = -\tfrac23$:

$$S_7 = \frac{729\left(1 – (-2/3)^7\right)}{1 + 2/3} = \frac{729\left(1 + \frac{128}{2187}\right)}{5/3} = \frac{729 + \frac{128}{3}}{5/3} = 463$$

Question 16

Find a GP for which the sum of the first two terms is $-4$ and the fifth term is $4$ times the third term.

Solution. The second condition first, since it involves only $r$:

$$ar^4 = 4ar^2 \quad\Longrightarrow\quad r^2 = 4 \quad\Longrightarrow\quad r = \pm2$$

Now use $a + ar = -4$, that is $a(1 + r) = -4$.

If $r = 2$: $3a = -4$, so $a = -\dfrac43$ and the GP is

$$-\frac43,\ -\frac83,\ -\frac{16}{3},\ \ldots$$

If $r = -2$: $a(-1) = -4$, so $a = 4$ and the GP is

$$4,\ -8,\ 16,\ -32,\ 64,\ \ldots$$

Question 17

If the $4$th, $10$th and $16$th terms of a GP are $x$, $y$ and $z$ respectively, prove that $x$, $y$, $z$ are in GP.

Solution.

$$x = ar^3, \qquad y = ar^9, \qquad z = ar^{15}$$

For $x, y, z$ to be in GP we need $y^2 = xz$:

$$y^2 = a^2 r^{18}, \qquad xz = (ar^3)(ar^{15}) = a^2r^{18}$$

They agree, so $x, y, z$ are in GP — with common ratio $\dfrac yx = r^6$. $\blacksquare$

The indices $4$, $10$, $16$ are in arithmetic progression, and terms of a GP at equally spaced positions always form a GP themselves.

Question 18

Find the sum to $n$ terms of the sequence $8, 88, 888, 8888, \ldots$

Solution. This is not a GP, but it becomes one after a rewrite. Take out the $8$ and multiply and divide by $9$:

$$8 + 88 + 888 + \cdots = \frac89\left(9 + 99 + 999 + \cdots\right)$$

Now each bracketed term is a power of $10$ minus $1$:

$$= \frac89\left[(10 – 1) + (10^2 – 1) + \cdots + (10^n – 1)\right]$$

$$= \frac89\left[\left(10 + 10^2 + \cdots + 10^n\right) – n\right]$$

The inner sum is a GP with $a = 10$, $r = 10$:

$$10 + 10^2 + \cdots + 10^n = \frac{10(10^n – 1)}{9}$$

Therefore

$$S_n = \frac89\left[\frac{10(10^n-1)}{9} – n\right] = \frac{80}{81}\left(10^n – 1\right) – \frac89 n$$

Question 19

Find the sum of the products of the corresponding terms of the sequences $2, 4, 8, 16, 32$ and $128, 32, 8, 2, \dfrac12$.

Solution. Pair the terms and multiply:

$$2 \times 128 = 256, \quad 4 \times 32 = 128, \quad 8 \times 8 = 64, \quad 16 \times 2 = 32, \quad 32 \times \frac12 = 16$$

These products themselves form a GP with $a = 256$ and $r = \tfrac12$ — which is question 20’s general result in a particular case. Summing:

$$S_5 = \frac{256\left(1 – (1/2)^5\right)}{1 – 1/2} = 512\left(1 – \frac{1}{32}\right) = 512 – 16 = 496$$

Question 20

Show that the products of the corresponding terms of the sequences $a, ar, ar^2, \ldots, ar^{n-1}$ and $\mathrm A, \mathrm{AR}, \mathrm{AR}^2, \ldots, \mathrm{AR}^{n-1}$ form a GP, and find the common ratio.

Solution. The $k$th terms are $ar^{k-1}$ and $\mathrm{AR}^{k-1}$, so their product is

$$P_k = a\mathrm A\,(r\mathrm R)^{k-1}$$

The ratio of consecutive products is

$$\frac{P_{k+1}}{P_k} = \frac{a\mathrm A (r\mathrm R)^{k}}{a\mathrm A (r\mathrm R)^{k-1}} = r\mathrm R$$

which is a constant independent of $k$. Hence the products form a GP with first term $a\mathrm A$ and common ratio $r\mathrm R$. $\blacksquare$

Question 21

Find four numbers forming a geometric progression in which the third term is greater than the first term by $9$, and the second term is greater than the $4$th by $18$.

Solution. Let the four numbers be $a, ar, ar^2, ar^3$. The conditions are

$$ar^2 – a = 9 \quad\Longrightarrow\quad a(r^2 – 1) = 9 \qquad (1)$$ $$ar – ar^3 = 18 \quad\Longrightarrow\quad -ar(r^2 – 1) = 18 \qquad (2)$$

Divide (2) by (1) — the common factor $a(r^2-1)$ cancels:

$$-r = \frac{18}{9} = 2 \quad\Longrightarrow\quad r = -2$$

Substituting into (1): $a(4 – 1) = 9$, so $a = 3$.

$$3,\ -6,\ 12,\ -24$$

Check: $12 – 3 = 9$ ✓ and $-6 – (-24) = 18$ ✓

Question 22

If the $p$th, $q$th and $r$th terms of a GP are $a$, $b$ and $c$ respectively, prove that $a^{q-r}\,b^{r-p}\,c^{p-q} = 1$.

Solution. Let the GP have first term $A$ and ratio $R$:

$$a = AR^{p-1}, \qquad b = AR^{q-1}, \qquad c = AR^{r-1}$$

Substituting into the left side and collecting powers of $A$ and $R$ separately:

$$a^{q-r}b^{r-p}c^{p-q} = A^{(q-r)+(r-p)+(p-q)}\;R^{(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)}$$

The exponent of $A$ telescopes to zero:

$$(q-r) + (r-p) + (p-q) = 0$$

The exponent of $R$: expand the products,

$$(pq – pr) + (qr – pq) + (rp – qr) – \left[(q-r) + (r-p) + (p-q)\right]$$

The first bracket cancels pairwise to $0$, and the second is $0$ as just shown. So the exponent of $R$ is $0$ too, and

$$a^{q-r}b^{r-p}c^{p-q} = A^0 R^0 = 1 \qquad \blacksquare$$

Question 23

If the first and the $n$th term of a GP are $a$ and $b$ respectively, and if $\mathrm P$ is the product of $n$ terms, prove that $\mathrm P^2 = (ab)^n$.

Solution. The terms are $a, ar, ar^2, \ldots, ar^{n-1}$, so their product is

$$\mathrm P = a^n\,r^{0+1+2+\cdots+(n-1)} = a^n\,r^{\frac{n(n-1)}{2}}$$

Squaring:

$$\mathrm P^2 = a^{2n}\,r^{n(n-1)}$$

On the other side, $b = ar^{n-1}$, so

$$(ab)^n = \left(a \cdot ar^{n-1}\right)^n = \left(a^2 r^{n-1}\right)^n = a^{2n}r^{n(n-1)}$$

The two agree. $\blacksquare$

Question 24

Show that the ratio of the sum of the first $n$ terms of a GP to the sum of terms from the $(n+1)$th to the $(2n)$th term is $\dfrac{1}{r^n}$.

Solution. The first sum is

$$S_n = \frac{a(r^n – 1)}{r – 1}$$

The terms from the $(n+1)$th to the $(2n)$th are $ar^n, ar^{n+1}, \ldots, ar^{2n-1}$ — that is, the first $n$ terms each multiplied by $r^n$. So their sum is

$$S’ = r^n \cdot S_n$$

Therefore

$$\frac{S_n}{S’} = \frac{S_n}{r^n S_n} = \frac{1}{r^n} \qquad \blacksquare$$

Seeing the second block as a scaled copy of the first avoids summing it from scratch.

Question 25

If $a$, $b$, $c$ and $d$ are in GP, show that

$$(a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (ab + bc + cd)^2$$

Solution. In a GP with ratio $r$, write $b = ar$, $c = ar^2$, $d = ar^3$. The key relations are

$$b^2 = ac, \qquad c^2 = bd, \qquad bc = ad$$

Right side:

$$(ab + bc + cd)^2 = \left(a^2r + a^2r^3 + a^2r^5\right)^2 = a^4r^2\left(1 + r^2 + r^4\right)^2$$

Left side:

$$a^2 + b^2 + c^2 = a^2\left(1 + r^2 + r^4\right)$$ $$b^2 + c^2 + d^2 = a^2r^2\left(1 + r^2 + r^4\right)$$

Multiplying:

$$a^4r^2\left(1 + r^2 + r^4\right)^2$$

The two sides agree. $\blacksquare$

Question 26

Insert two numbers between $3$ and $81$ so that the resulting sequence is a GP.

Solution. The sequence is $3, G_1, G_2, 81$ — four terms, so $81$ is the fourth term:

$$3r^3 = 81 \quad\Longrightarrow\quad r^3 = 27 \quad\Longrightarrow\quad r = 3$$

(A cube root of a positive number is unique among reals, so there is no $\pm$ here — unlike the even powers in questions 15 and 16.)

$$G_1 = 3 \times 3 = 9, \qquad G_2 = 9 \times 3 = 27$$

Question 27

Find the value of $n$ so that $\dfrac{a^{n+1} + b^{n+1}}{a^n + b^n}$ may be the geometric mean between $a$ and $b$.

Solution. The geometric mean is $\sqrt{ab} = a^{1/2}b^{1/2}$, so set

$$\frac{a^{n+1} + b^{n+1}}{a^n + b^n} = a^{1/2}b^{1/2}$$

Cross-multiplying:

$$a^{n+1} + b^{n+1} = a^{n + \frac12}b^{\frac12} + a^{\frac12}b^{n+\frac12}$$

Group the terms so each side has a common factor:

$$a^{n+1} – a^{n+\frac12}b^{\frac12} = a^{\frac12}b^{n+\frac12} – b^{n+1}$$

$$a^{n+\frac12}\left(a^{\frac12} – b^{\frac12}\right) = b^{n+\frac12}\left(a^{\frac12} – b^{\frac12}\right)$$

Since $a \ne b$, divide by $\left(a^{1/2} – b^{1/2}\right)$:

$$a^{n+\frac12} = b^{n+\frac12} \quad\Longrightarrow\quad \left(\frac ab\right)^{n + \frac12} = 1$$

As $a \ne b$, the exponent must vanish:

$$n + \frac12 = 0 \quad\Longrightarrow\quad n = -\frac12$$

Question 28

The sum of two numbers is $6$ times their geometric mean. Show that the numbers are in the ratio $\left(3 + 2\sqrt2\right) : \left(3 – 2\sqrt2\right)$.

Solution. Let the numbers be $a$ and $b$. The condition is

$$a + b = 6\sqrt{ab}$$

Use the standard componendo-and-dividendo device. Adding and subtracting $2\sqrt{ab}$:

$$a + b + 2\sqrt{ab} = 6\sqrt{ab} + 2\sqrt{ab} = 8\sqrt{ab} \quad\Longrightarrow\quad \left(\sqrt a + \sqrt b\right)^2 = 8\sqrt{ab}$$

$$a + b – 2\sqrt{ab} = 6\sqrt{ab} – 2\sqrt{ab} = 4\sqrt{ab} \quad\Longrightarrow\quad \left(\sqrt a – \sqrt b\right)^2 = 4\sqrt{ab}$$

Dividing and taking square roots:

$$\frac{\left(\sqrt a + \sqrt b\right)^2}{\left(\sqrt a – \sqrt b\right)^2} = \frac{8\sqrt{ab}}{4\sqrt{ab}} = 2 \quad\Longrightarrow\quad \frac{\sqrt a + \sqrt b}{\sqrt a – \sqrt b} = \sqrt2$$

By componendo and dividendo,

$$\frac{\sqrt a}{\sqrt b} = \frac{\sqrt2 + 1}{\sqrt2 – 1}$$

Squaring both sides:

$$\frac ab = \frac{(\sqrt2+1)^2}{(\sqrt2-1)^2} = \frac{3 + 2\sqrt2}{3 – 2\sqrt2}$$

$$a : b = \left(3 + 2\sqrt2\right) : \left(3 – 2\sqrt2\right) \qquad \blacksquare$$

Question 29

If $\mathrm A$ and $\mathrm G$ are the AM and GM respectively between two positive numbers, prove that the numbers are $\mathrm A \pm \sqrt{(\mathrm A + \mathrm G)(\mathrm A – \mathrm G)}$.

Solution. Let the numbers be $a$ and $b$. By definition

$$\mathrm A = \frac{a+b}{2} \quad\Longrightarrow\quad a + b = 2\mathrm A$$ $$\mathrm G = \sqrt{ab} \quad\Longrightarrow\quad ab = \mathrm G^2$$

The two numbers are therefore the roots of

$$x^2 – 2\mathrm A x + \mathrm G^2 = 0$$

Solving by the quadratic formula:

$$x = \frac{2\mathrm A \pm \sqrt{4\mathrm A^2 – 4\mathrm G^2}}{2} = \mathrm A \pm \sqrt{\mathrm A^2 – \mathrm G^2}$$

Factorising the difference of two squares:

$$x = \mathrm A \pm \sqrt{(\mathrm A + \mathrm G)(\mathrm A – \mathrm G)} \qquad \blacksquare$$

(The AM–GM inequality guarantees $\mathrm A \ge \mathrm G$, so the radicand is non-negative and both roots are real.)

Question 30

The number of bacteria in a certain culture doubles every hour. If there were $30$ bacteria present in the culture originally, how many bacteria will be present at the end of the $2$nd hour, $4$th hour and $n$th hour?

Solution. Doubling every hour is a GP with $a = 30$ and $r = 2$. After $n$ hours the count is $30 \times 2^n$ — note the exponent is $n$, not $n-1$, because the original count corresponds to $n = 0$.

$$\text{2nd hour: } 30 \times 2^2 = 120$$ $$\text{4th hour: } 30 \times 2^4 = 480$$ $$n\text{th hour: } 30 \times 2^n$$

Question 31

What will Rs $500$ amount to in $10$ years after its deposit in a bank which pays an annual interest rate of $10\%$ compounded annually?

Solution. Compounding annually multiplies the amount by $1 + \tfrac{10}{100} = 1.1$ each year, so the amounts form a GP:

$$500,\ 500(1.1),\ 500(1.1)^2,\ \ldots$$

After $10$ years:

$$\text{Rs } 500(1.1)^{10}$$

(Numerically this is about Rs $1296.87$, but NCERT leaves the answer in exact form.)

Question 32

If the AM and GM of the roots of a quadratic equation are $8$ and $5$ respectively, obtain the quadratic equation.

Solution. Let the roots be $\alpha$ and $\beta$.

$$\text{AM} = \frac{\alpha + \beta}{2} = 8 \quad\Longrightarrow\quad \alpha + \beta = 16$$

$$\text{GM} = \sqrt{\alpha\beta} = 5 \quad\Longrightarrow\quad \alpha\beta = 25$$

A quadratic with these roots is

$$x^2 – (\text{sum})x + (\text{product}) = 0$$

$$x^2 – 16x + 25 = 0$$

Common mistakes

• Question 4, dividing by $r$ once too few times. From $-3r^3 = 9r^2$, cancelling $r^2$ leaves $-3r = 9$, so $r = -3$. Cancelling $r^3$ instead loses the equation.
• Question 15, taking only the positive sixth root. $r^6 = \left(\tfrac23\right)^6$ has two real solutions, and they give genuinely different sums, $2059$ and $463$. Question 26, by contrast, involves a cube root and has only one.
• Question 12, parametrising as $a, ar, ar^2$. The product then becomes $a^3r^3$, entangling both unknowns. With $\tfrac ar, a, ar$ the product is $a^3$ and $a$ falls out at once.
• Question 18, treating $8, 88, 888$ as a GP. It is not — the ratios are $11, 10.09\ldots$ The trick is $\tfrac89 \times (9 + 99 + 999 + \cdots)$, turning each term into $10^k – 1$.
• Question 21, forgetting that the ratio of the two equations kills $a$. Dividing (2) by (1) removes both $a$ and $(r^2-1)$ in one step; solving them simultaneously by substitution is much longer.
• Question 24, summing the second block from scratch. It is the first block multiplied by $r^n$, which makes the ratio immediate.
• Question 27, forgetting that $a \ne b$ is needed. The division by $\left(\sqrt a – \sqrt b\right)$ is legitimate only then, and it is that step which forces $n + \tfrac12 = 0$.
• Question 30, using $30 \times 2^{n-1}$. The initial count is at $n = 0$ hours, so after $n$ hours the population is $30 \times 2^n$. Check against the second hour: $120$, not $60$.

Practise next

• Miscellaneous Exercise on Chapter 8 — sums of special series, instalment problems, and mixed AP–GP questions built on everything here.
• Exercise 8.1 — worth revising if generating terms from a rule still feels slow.