Sequences and Series — NCERT Solutions

Given: $S_n = 2n - 3$

Concept: The $n^{\text{th}}$ term of a sequence is $a_n = S_n - S_{n-1}$.

Working:
$$a_n = S_n - S_{n-1} = (2n-3) - (2(n-1)-3) = (2n-3)-(2n-5) = 2$$

Since every term equals 2 (a constant), the common difference $d = a_n - a_{n-1} = 2$.

Answer: (b) 2

Given: Integers from 100 to 500 divisible by 5.

Concept: These form an A.P.: $100, 105, 110, \ldots, 500$ with $a = 100$, $d = 5$, $l = 500$.

Working:
$$l = a + (n-1)d$$
$$500 = 100 + (n-1)\times 5$$
$$400 = 5(n-1)$$
$$n-1 = 80 \Rightarrow n = 81$$

Answer: (b) 81

Given: Two-digit numbers divisible by 6.

Concept: These form an A.P.: $12, 18, 24, \ldots, 96$ with $a = 12$, $d = 6$, $l = 96$.

Working:
$$l = a + (n-1)d$$
$$96 = 12 + (n-1)\times 6$$
$$84 = 6(n-1)$$
$$n-1 = 14 \Rightarrow n = 15$$

Answer: (c) 15

Given: $a_4 = 4$

Concept: For an A.P., $S_n = \dfrac{n}{2}(a + l)$. Also, the middle term of a 7-term A.P. is $a_4$, and $S_7 = 7 \times a_4$.

Working:
$$S_7 = \frac{7}{2}[2a + 6d] = 7(a + 3d) = 7 \times a_4 = 7 \times 4 = 28$$

Answer: (a) 28

Given: A.P. with $a = -3$, $d = -4$.

Concept: $a_n = a + (n-1)d = -3 + (n-1)(-4) = -3 - 4n + 4 = 1 - 4n$.

For a number $k$ to be a term: $k = 1 - 4n \Rightarrow n = \dfrac{1-k}{4}$ must be a positive integer.

Checking each option:

• (a) $k = -500$: $n = \dfrac{1-(-500)}{4} = \dfrac{501}{4} = 125.25$ — not an integer, so $-500$ is NOT a term.
• (b) $k = -399$: $n = \dfrac{400}{4} = 100$ — integer, so it IS a term.
• (c) $k = -503$: $n = \dfrac{504}{4} = 126$ — integer, so it IS a term.
• (d) $k = -51$: $n = \dfrac{52}{4} = 13$ — integer, so it IS a term.

Answer: (a) $-500$

Given: An A.P. with first term $a$ and common difference $d$: $a,\ a+d,\ a+2d,\ \ldots$

New sequence (each term doubled): $2a,\ 2(a+d),\ 2(a+2d),\ \ldots$

Check for A.P.:
Difference between consecutive terms:
$$2(a+d) - 2a = 2d$$
$$2(a+2d) - 2(a+d) = 2d$$

The difference is constant $= 2d$.

Conclusion: Yes, the new sequence is an A.P. with common difference $2d$ (twice the common difference of the original A.P.).

Given: A.P. with $a = 20$, $d = -4$, $l = -176$.

Step 1: Find number of terms.
$$l = a + (n-1)d$$
$$-176 = 20 + (n-1)(-4)$$
$$-196 = -4(n-1)$$
$$n-1 = 49 \Rightarrow n = 50$$

Step 2: Find middle terms.
Since $n = 50$ (even), there are two middle terms: $a_{25}$ and $a_{26}$.

$$a_{25} = 20 + 24(-4) = 20 - 96 = -76$$
$$a_{26} = 20 + 25(-4) = 20 - 100 = -80$$

Answer: The middle terms are $a_{25} = -76$ and $a_{26} = -80$.

Given: $a_4 : a_7 = 2 : 3$

Let first term $= a$, common difference $= d$.

$$\frac{a + 3d}{a + 6d} = \frac{2}{3}$$
$$3(a + 3d) = 2(a + 6d)$$
$$3a + 9d = 2a + 12d$$
$$a = 3d$$

Now find $a_5 : a_8$:
$$a_5 = a + 4d = 3d + 4d = 7d$$
$$a_8 = a + 7d = 3d + 7d = 10d$$
$$a_5 : a_8 = 7d : 10d = \boxed{7 : 10}$$

(Note: The answer key states $a_6:a_8 = 4:5$; the question asks for $a_5:a_8 = 7:10$.)

Given: Integers from 100 to 500 divisible by both 2 and 3, i.e., divisible by LCM$(2,3) = 6$.

A.P.: $102, 108, 114, \ldots, 498$ with $a = 102$, $d = 6$, $l = 498$.

Step 1: Find $n$.
$$498 = 102 + (n-1)\times 6$$
$$396 = 6(n-1)$$
$$n = 67$$

Step 2: Find sum.
$$S = \frac{n}{2}(a + l) = \frac{67}{2}(102 + 498) = \frac{67}{2} \times 600 = 67 \times 300 = \boxed{20100}$$

Given: $a_n = 2n + 1$

First term: $a_1 = 2(1)+1 = 3$

Common difference: $d = a_2 - a_1 = 5 - 3 = 2$

Sum of 20 terms:
$$S_{20} = \frac{20}{2}[2a + 19d] = 10[2(3) + 19(2)] = 10[6 + 38] = 10 \times 44 = \boxed{440}$$

Given: $n$ A.M.s are inserted between 1 and 31. So the full sequence is:
$$1, A_1, A_2, \ldots, A_n, 31$$
Total terms $= n + 2$, first term $a = 1$, last term $= 31$.

Common difference:
$$d = \frac{31 - 1}{n + 1} = \frac{30}{n+1}$$

The $k^{\text{th}}$ arithmetic mean: $A_k = 1 + kd$

$$A_7 = 1 + 7d, \quad A_{n-1} = 1 + (n-1)d$$

Given ratio:
$$\frac{A_7}{A_{n-1}} = \frac{5}{9}$$
$$\frac{1 + 7d}{1 + (n-1)d} = \frac{5}{9}$$

Substitute $d = \dfrac{30}{n+1}$:
$$\frac{1 + \dfrac{210}{n+1}}{1 + \dfrac{30(n-1)}{n+1}} = \frac{5}{9}$$

Multiply numerator and denominator by $(n+1)$:
$$\frac{(n+1) + 210}{(n+1) + 30(n-1)} = \frac{5}{9}$$
$$\frac{n + 211}{n + 1 + 30n - 30} = \frac{5}{9}$$
$$\frac{n + 211}{31n - 29} = \frac{5}{9}$$
$$9(n + 211) = 5(31n - 29)$$
$$9n + 1899 = 155n - 145$$
$$2044 = 146n$$
$$n = 14$$

Common difference: $d = \dfrac{30}{15} = 2$

Resulting A.P.: $1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31$

Answer: $n = 14$; the A.P. is $1, 3, 5, 7, \ldots, 31$ with $d = 2$.

(a) $4 + 7 + 10 + \dots$ to 100 terms

$a = 4$, $d = 3$, $n = 100$
$$S_{100} = \frac{100}{2}[2(4) + 99(3)] = 50[8 + 297] = 50 \times 305 = \boxed{15250}$$

(b) $1 + \dfrac{4}{3} + \dfrac{5}{3} + 2 + \dots$ to 19 terms

$a = 1$, $d = \dfrac{4}{3} - 1 = \dfrac{1}{3}$, $n = 19$
$$S_{19} = \frac{19}{2}\left[2(1) + 18\left(\frac{1}{3}\right)\right] = \frac{19}{2}\left[2 + 6\right] = \frac{19}{2} \times 8 = \boxed{76}$$

(c) $0.5 + 0.51 + 0.52 + \dots$ to 1000 terms

$a = 0.5$, $d = 0.01$, $n = 1000$
$$S_{1000} = \frac{1000}{2}[2(0.5) + 999(0.01)] = 500[1 + 9.99] = 500 \times 10.99 = \boxed{5495}$$

Given: $S_n = 72$, $a = 17$, $d = -2$

Formula:
$$S_n = \frac{n}{2}[2a + (n-1)d]$$
$$72 = \frac{n}{2}[2(17) + (n-1)(-2)]$$
$$144 = n[34 - 2n + 2]$$
$$144 = n[36 - 2n]$$
$$144 = 36n - 2n^2$$
$$2n^2 - 36n + 144 = 0$$
$$n^2 - 18n + 72 = 0$$
$$(n - 6)(n - 12) = 0$$
$$n = 6 \quad \text{or} \quad n = 12$$

Verification:

• $S_6 = \frac{6}{2}[34 + 5(-2)] = 3 \times 24 = 72$ ✓
• $S_{12} = \frac{12}{2}[34 + 11(-2)] = 6 \times 12 = 72$ ✓

Answer: $n = 6$ or $n = 12$.

Given: $S_p = S_q$, where $S_n = \dfrac{n}{2}[2a + (n-1)d]$.

To prove: $S_{p+q} = 0$

Proof:
$$S_p = S_q$$
$$\frac{p}{2}[2a + (p-1)d] = \frac{q}{2}[2a + (q-1)d]$$
$$p[2a + (p-1)d] = q[2a + (q-1)d]$$
$$2ap + p(p-1)d = 2aq + q(q-1)d$$
$$2a(p - q) + d[p(p-1) - q(q-1)] = 0$$
$$2a(p - q) + d[p^2 - p - q^2 + q] = 0$$
$$2a(p - q) + d[(p^2 - q^2) - (p - q)] = 0$$
$$2a(p - q) + d(p - q)(p + q - 1) = 0$$

Since $p \neq q$, divide by $(p - q)$:
$$2a + d(p + q - 1) = 0 \quad \ldots (1)$$

Now compute $S_{p+q}$:
$$S_{p+q} = \frac{p+q}{2}[2a + (p+q-1)d]$$

From (1): $2a + (p+q-1)d = 0$

$$\therefore S_{p+q} = \frac{p+q}{2} \times 0 = 0$$

Hence proved. $\blacksquare$

Given: A.P.: $19,\ \dfrac{91}{5},\ \dfrac{87}{5},\ \ldots$

$a = 19 = \dfrac{95}{5}$, $d = \dfrac{91}{5} - \dfrac{95}{5} = -\dfrac{4}{5}$

General term:
$$a_n = a + (n-1)d = \frac{95}{5} + (n-1)\left(-\frac{4}{5}\right) = \frac{95 - 4(n-1)}{5} = \frac{99 - 4n}{5}$$

For first negative term: $a_n < 0$
$$\frac{99 - 4n}{5} < 0$$
$$99 - 4n < 0$$
$$n > \frac{99}{4} = 24.75$$

So the first negative term is at $n = 25$.

$$a_{25} = \frac{99 - 4(25)}{5} = \frac{99 - 100}{5} = \frac{-1}{5}$$

Answer: The first negative term is the $25^{\text{th}}$ term $= -\dfrac{1}{5}$.

Given: Let first term of A.P. be $A$ and common difference be $D$.

$$a = A + (p-1)D, \quad b = A + (q-1)D, \quad c = A + (r-1)D$$

Compute differences:
$$a - b = (p-1)D - (q-1)D = (p-q)D$$
$$b - c = (q-r)D$$
$$c - a = (r-p)D$$

LHS:
$$(a-b)r + (b-c)p + (c-a)q$$
$$= D[(p-q)r + (q-r)p + (r-p)q]$$
$$= D[pr - qr + pq - pr + qr - pq]$$
$$= D[0]$$
$$= 0 = \text{RHS}$$

Hence proved. $\blacksquare$

Given: $a = 5$, $d = 3$, $l = 95$.

Step 1: Find $n$.
$$95 = 5 + (n-1)(3) \Rightarrow 90 = 3(n-1) \Rightarrow n = 31$$

Step 2: Middle term.
Since $n = 31$ (odd), there is one middle term: $a_{16}$.

$$a_{16} = 5 + 15 \times 3 = 5 + 45 = \boxed{50}$$

Answer: The middle term is the $16^{\text{th}}$ term $= 50$.

Given: $a_p = \dfrac{1}{q}$ and $a_q = \dfrac{1}{p}$.

Step 1: Find $a$ and $d$.
$$a + (p-1)d = \frac{1}{q} \quad \ldots (1)$$
$$a + (q-1)d = \frac{1}{p} \quad \ldots (2)$$

Subtracting (2) from (1):
$$(p-q)d = \frac{1}{q} - \frac{1}{p} = \frac{p-q}{pq}$$
$$d = \frac{1}{pq}$$

From (1): $a = \dfrac{1}{q} - (p-1)\dfrac{1}{pq} = \dfrac{p - (p-1)}{pq} = \dfrac{1}{pq}$

Step 2: Sum of $pq$ terms.
$$S_{pq} = \frac{pq}{2}[2a + (pq-1)d]$$
$$= \frac{pq}{2}\left[\frac{2}{pq} + (pq-1)\cdot\frac{1}{pq}\right]$$
$$= \frac{pq}{2} \cdot \frac{2 + pq - 1}{pq}$$
$$= \frac{pq+1}{2} = \frac{1}{2}(pq+1)$$

Hence proved. $\blacksquare$

Given: $\dfrac{S_n}{S_n'} = \dfrac{5n+4}{9n+6}$

Concept: The ratio of $k^{\text{th}}$ terms equals the ratio of sums when $n = 2k - 1$.

For the $18^{\text{th}}$ term, put $n = 2(18) - 1 = 35$:

$$\frac{a_{18}}{a_{18}'} = \frac{5(35)+4}{9(35)+6} = \frac{175+4}{315+6} = \frac{179}{321}$$

Answer: The ratio of their $18^{\text{th}}$ terms is $\boxed{179 : 321}$.

Given: $a = 1000$, $d = -50$.

For zero patients: $a_n = 0$
$$1000 + (n-1)(-50) = 0$$
$$1000 = 50(n-1)$$
$$n - 1 = 20 \Rightarrow n = 21$$

Answer: Yes, on the 21st day from the day when there were 1000 patients, there would be no COVID patients in the OPD.

Given: G.P. with $a = 4$, $r = \dfrac{12}{4} = 3$, find $a_5$.

$$a_5 = ar^{5-1} = 4 \times 3^4 = 4 \times 81 = \boxed{324}$$

Given: $a = 3$, $r = \dfrac{-1}{3}$.

4th term:
$$a_4 = ar^3 = 3 \times \left(-\frac{1}{3}\right)^3 = 3 \times \left(-\frac{1}{27}\right) = -\frac{1}{9}$$

nth term:
$$a_n = ar^{n-1} = 3 \times \left(-\frac{1}{3}\right)^{n-1} = 3 \times \frac{(-1)^{n-1}}{3^{n-1}} = \frac{(-1)^{n-1}}{3^{n-2}}$$

Answer: $a_4 = -\dfrac{1}{9}$; $a_n = \dfrac{(-1)^{n-1}}{3^{n-2}}$

Given: $a = 5$, $r = 2$, $a_n = 5120$.

$$5 \times 2^{n-1} = 5120$$
$$2^{n-1} = 1024 = 2^{10}$$
$$n - 1 = 10 \Rightarrow n = 11$$

Answer: 5120 is the 11th term.

Given: $a = 2$, $r = \dfrac{2\sqrt{2}}{2} = \sqrt{2}$, $a_n = 128$.

$$2 \times (\sqrt{2})^{n-1} = 128$$
$$(\sqrt{2})^{n-1} = 64 = 2^6 = (\sqrt{2})^{12}$$
$$n - 1 = 12 \Rightarrow n = 13$$

Answer: 128 is the 13th term.

Given: $a = 2$, $r = \dfrac{1}{2}$, $a_n = \dfrac{1}{128}$.

$$2 \times \left(\frac{1}{2}\right)^{n-1} = \frac{1}{128}$$
$$\left(\frac{1}{2}\right)^{n-1} = \frac{1}{256} = \left(\frac{1}{2}\right)^8$$
$$n - 1 = 8 \Rightarrow n = 9$$

Answer: $\dfrac{1}{128}$ is the 9th term.

Given: $a = \sqrt{3}$, $r = \dfrac{3}{\sqrt{3}} = \sqrt{3}$, $n = 6$.

$$S_6 = \frac{a(r^6 - 1)}{r - 1} = \frac{\sqrt{3}\left[(\sqrt{3})^6 - 1\right]}{\sqrt{3} - 1} = \frac{\sqrt{3}(27 - 1)}{\sqrt{3} - 1} = \frac{26\sqrt{3}}{\sqrt{3} - 1}$$

Rationalise:
$$= \frac{26\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{26\sqrt{3}(\sqrt{3}+1)}{2} = 13\sqrt{3}(\sqrt{3}+1) = 13(3 + \sqrt{3}) = 39 + 13\sqrt{3}$$

Answer: $S_6 = 39 + 13\sqrt{3}$

Given: $a = 0.15$, $r = 0.1$, $n = 20$.

$$S_{20} = \frac{a(1 - r^{20})}{1 - r} = \frac{0.15(1 - (0.1)^{20})}{1 - 0.1} = \frac{0.15}{0.9}\left[1 - (0.1)^{20}\right] = \frac{1}{6}\left[1 - (0.1)^{20}\right]$$

Answer: $S_{20} = \dfrac{1}{6}\left[1 - (0.1)^{20}\right]$

$$\sum_{k=1}^{10}(3 + 2^k) = \sum_{k=1}^{10} 3 + \sum_{k=1}^{10} 2^k$$

$$= 3 \times 10 + (2^1 + 2^2 + \ldots + 2^{10})$$

$$= 30 + \frac{2(2^{10}-1)}{2-1} = 30 + 2(1024 - 1) = 30 + 2046 = 2076$$

Alternatively expressed: $28 + 2^{11} = 28 + 2048 = 2076$.

Answer: $\displaystyle\sum_{k=1}^{10}(3+2^k) = 2076$

Given: $a_1 + a_2 = 36$ and $a_1 \cdot a_3 = 9a_2$.

Let first term $= a$, common ratio $= r$.

Condition 2: $a \cdot ar^2 = 9 \cdot ar$
$$a^2r^2 = 9ar \Rightarrow ar = 9 \Rightarrow a_2 = 9$$

Condition 1: $a + 9 = 36 \Rightarrow a = 27$

$$r = \frac{9}{27} = \frac{1}{3}$$

Sum of 8 terms:
$$S_8 = \frac{27\left(1 - \left(\frac{1}{3}\right)^8\right)}{1 - \frac{1}{3}} = \frac{27\left(1 - \frac{1}{6561}\right)}{\frac{2}{3}} = \frac{27 \times \frac{6560}{6561}}{\frac{2}{3}} = \frac{27 \times 6560 \times 3}{6561 \times 2} = \frac{6560}{162} = \frac{3280}{81}$$

Answer: $S_8 = \dfrac{3280}{81}$

Given: $S_n = 7 + 77 + 777 + \ldots$ to $n$ terms.

$$S_n = 7(1 + 11 + 111 + \ldots \text{ to } n \text{ terms})$$
$$= \frac{7}{9}(9 + 99 + 999 + \ldots)$$
$$= \frac{7}{9}[(10-1) + (10^2-1) + (10^3-1) + \ldots + (10^n-1)]$$
$$= \frac{7}{9}\left[(10 + 10^2 + \ldots + 10^n) - n\right]$$
$$= \frac{7}{9}\left[\frac{10(10^n - 1)}{9} - n\right]$$
$$= \frac{70(10^n - 1)}{81} - \frac{7n}{9}$$

Answer: $S_n = \dfrac{70(10^n - 1)}{81} - \dfrac{7n}{9}$

Let the three terms be $\dfrac{a}{r}, a, ar$.

Product: $\dfrac{a}{r} \cdot a \cdot ar = a^3 = 1 \Rightarrow a = 1$

Sum: $\dfrac{1}{r} + 1 + r = \dfrac{39}{10}$

$$\frac{1 + r + r^2}{r} = \frac{39}{10}$$
$$10(1 + r + r^2) = 39r$$
$$10r^2 - 29r + 10 = 0$$
$$(2r - 5)(5r - 2) = 0$$
$$r = \frac{5}{2} \quad \text{or} \quad r = \frac{2}{5}$$

Terms:

• If $r = \dfrac{5}{2}$: terms are $\dfrac{2}{5}, 1, \dfrac{5}{2}$
• If $r = \dfrac{2}{5}$: terms are $\dfrac{5}{2}, 1, \dfrac{2}{5}$

Answer: $r = \dfrac{5}{2}$ or $\dfrac{2}{5}$; terms are $\dfrac{2}{5}, 1, \dfrac{5}{2}$ (or in reverse order).

Let the four terms be $a, ar, ar^2, ar^3$.

Condition 1: $ar^2 - a = 9 \Rightarrow a(r^2 - 1) = 9 \quad \ldots(1)$

Condition 2: $ar - ar^3 = 18 \Rightarrow ar(1 - r^2) = 18 \quad \ldots(2)$

Divide (2) by (1):
$$\frac{ar(1-r^2)}{a(r^2-1)} = \frac{18}{9}$$
$$\frac{-ar(r^2-1)}{a(r^2-1)} = 2$$
$$-r = 2 \Rightarrow r = -2$$

From (1): $a(4-1) = 9 \Rightarrow a = 3$

Four terms: $3,\ 3(-2),\ 3(-2)^2,\ 3(-2)^3 = 3, -6, 12, -24$

Answer: The four numbers are $3, -6, 12, -24$.

Given: 6 G.M.s between 27 and $\dfrac{1}{81}$. Total terms $= 8$.

So: $a_1 = 27$, $a_8 = \dfrac{1}{81}$, $r = ?$

$$a_8 = a_1 r^7 \Rightarrow \frac{1}{81} = 27 r^7$$
$$r^7 = \frac{1}{81 \times 27} = \frac{1}{2187} = \frac{1}{3^7} = \left(\frac{1}{3}\right)^7$$
$$r = \frac{1}{3}$$

Six G.M.s:
$$G_1 = 27 \times \frac{1}{3} = 9, \quad G_2 = 3, \quad G_3 = 1, \quad G_4 = \frac{1}{3}, \quad G_5 = \frac{1}{9}, \quad G_6 = \frac{1}{27}$$

Answer: The 6 geometric means are $9, 3, 1, \dfrac{1}{3}, \dfrac{1}{9}, \dfrac{1}{27}$.

Given: $A = 2G$, where $A = \dfrac{a+b}{2}$ and $G = \sqrt{ab}$.

$$\frac{a+b}{2} = 2\sqrt{ab}$$
$$a + b = 4\sqrt{ab} \quad \ldots (1)$$

Let $\dfrac{a}{b} = k$. Then:
$$\frac{a+b}{\sqrt{ab}} = 4$$
$$\frac{\sqrt{a/b} + \sqrt{b/a}}{1} = 4 \Rightarrow \sqrt{k} + \frac{1}{\sqrt{k}} = 4$$

Let $\sqrt{k} = t$:
$$t + \frac{1}{t} = 4 \Rightarrow t^2 - 4t + 1 = 0$$
$$t = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$$

Since $a > b$, $k > 1$, so $t = 2 + \sqrt{3}$.

$$\sqrt{\frac{a}{b}} = 2 + \sqrt{3} \Rightarrow \frac{a}{b} = (2+\sqrt{3})^2 = 7 + 4\sqrt{3}$$

Also: $\dfrac{a}{b} = \dfrac{(2+\sqrt{3})^2}{1} = \dfrac{(2+\sqrt{3})(2+\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} \cdot (2-\sqrt{3})$

More directly: $\dfrac{a}{b} = \dfrac{2+\sqrt{3}}{2-\sqrt{3}}$ (rationalising: $(2+\sqrt{3})(2+\sqrt{3}) = 7+4\sqrt{3}$ and $(2-\sqrt{3})(2+\sqrt{3})=1$, so $\dfrac{2+\sqrt{3}}{2-\sqrt{3}} = (2+\sqrt{3})^2 = 7+4\sqrt{3}$ ✓).

$$\therefore a : b = (2+\sqrt{3}) : (2-\sqrt{3}) \quad \blacksquare$$

Given: $a, b, c, d$ are in G.P. with common ratio $r$, so $b = ar$, $c = ar^2$, $d = ar^3$.

(i) Let $P = a^2+b^2$, $Q = b^2+c^2$, $R = c^2+d^2$.

$$P = a^2 + a^2r^2 = a^2(1+r^2)$$
$$Q = a^2r^2 + a^2r^4 = a^2r^2(1+r^2)$$
$$R = a^2r^4 + a^2r^6 = a^2r^4(1+r^2)$$

$$\frac{Q}{P} = r^2, \quad \frac{R}{Q} = r^2$$

Since the ratio is constant, $P, Q, R$ are in G.P. $\blacksquare$

(ii) From (i), $P, Q, R$ are in G.P., so $Q^2 = PR$.

$$\frac{1}{Q^2} = \frac{1}{PR} \Rightarrow \frac{1}{Q} \cdot \frac{1}{Q} = \frac{1}{P} \cdot \frac{1}{R}$$

This means $\dfrac{1}{P}, \dfrac{1}{Q}, \dfrac{1}{R}$ are in G.P. $\blacksquare$