Chapter 6: Sequences and Series

Let the two numbers be $a$ and $b$. We are given $a = 12$ and their Arithmetic Mean (A.M.) is 17.

Using the Arithmetic Mean formula: $$\text{A.M.} = \frac{a + b}{2}$$

$$17 = \frac{12 + b}{2} \implies 12 + b = 34 \implies b = 34 - 12 = 22$$

Therefore, the other number is 22. Correct option is **(b)**.

Let the 7 arithmetic means inserted between $a = 2$ and $b = 34$ be $A_1, A_2, \dots, A_7$.

The common difference $d$ when inserting $n$ arithmetic means is: $$d = \frac{b - a}{n + 1} = \frac{34 - 2}{7 + 1} = \frac{32}{8} = 4$$

The $k^{\text{th}}$ arithmetic mean is $A_k = a + k d$.

For the $4^{\text{th}}$ arithmetic mean: $$A_4 = a + 4d = 2 + 4(4) = 2 + 16 = 18$$

Therefore, the $4^{\text{th}}$ arithmetic mean is 18. Correct option is **(c)**.

For the given Geometric Progression: First term $a = 3$, Common ratio $r = \frac{6}{3} = 2$.

The $n^{\text{th}}$ term of a G.P. is given by: $$T_n = a r^{n-1}$$

$$T_5 = 3 \cdot (2)^{5-1} = 3 \cdot 2^4 = 3 \cdot 16 = 48$$

Therefore, the $5^{\text{th}}$ term is 48. Correct option is **(b)**.

Given G.P.: $a = 2$, common ratio $r = \frac{8}{2} = 4$.

Let the $n^{\text{th}}$ term be $T_n = 131072$.

$$T_n = a r^{n-1} \implies 2 \cdot 4^{n-1} = 131072$$

$$4^{n-1} = \frac{131072}{2} = 65536$$

Since $4^8 = (2^2)^8 = 2^{16} = 65536$, we have: $$n - 1 = 8 \implies n = 9$$

Thus, the $9^{\text{th}}$ term is 131072. Correct option is **(b)**.

Let the first 5 terms of the G.P. be $\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2$.

Here, the $3^{\text{rd}}$ (middle) term is $a = 6$.

The product of the first 5 terms is: $$\text{Product} = \left(\frac{a}{r^2}\right) \cdot \left(\frac{a}{r}\right) \cdot a \cdot (ar) \cdot (ar^2) = a^5$$

$$\text{Product} = 6^5$$

Therefore, the product is $6^5$. Correct option is **(b)**.

Here $a = 1$, common ratio $r = \frac{1}{3} < 1$, and $n = 6$.

Sum formula: $$S_n = \frac{a(1 - r^n)}{1 - r}$$

$$S_6 = \frac{1 \cdot \left(1 - \left(\frac{1}{3}\right)^6\right)}{1 - \frac{1}{3}} = \frac{1 - \frac{1}{729}}{\frac{2}{3}} = \frac{\frac{728}{729}}{\frac{2}{3}} = \frac{728}{729} \times \frac{3}{2} = \frac{364}{243}$$

Therefore, $S_6 = \frac{364}{243}$. Correct option is **(a)**.

Given $a = 3$, common ratio $r = 3$, and $S_n = 120$.

$$S_n = \frac{a(r^n - 1)}{r - 1} \implies 120 = \frac{3(3^n - 1)}{3 - 1}$$

$$120 = \frac{3(3^n - 1)}{2} \implies 3^n - 1 = \frac{120 \times 2}{3} = 80$$

$$3^n = 81 = 3^4 \implies n = 4$$

Therefore, 4 terms are required. Correct option is **(b)**.

Let $G_1, G_2, G_3$ be three geometric means inserted between $a = 1$ and $b = 256$.

The common ratio $r$ when inserting $n$ geometric means is: $$r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}} = \left(\frac{256}{1}\right)^{\frac{1}{3+1}} = (256)^{1/4} = 4$$

• $G_1 = a r = 1 \times 4 = 4$

• $G_2 = a r^2 = 1 \times 4^2 = 16$

• $G_3 = a r^3 = 1 \times 4^3 = 64$

Therefore, the three geometric means are 4, 16, 64. Correct option is **(d)**.

Let the infinite G.P. be $a, ar, ar^2, \dots$ with $|r| < 1$.

1. Sum of infinite G.P.: $$S_\infty = \frac{a}{1 - r} = 3 \implies a = 3(1 - r) \quad \text{--- (1)}$$

2. Sum of squares of terms ($a^2, a^2r^2, a^2r^4, \dots$): $$S_{\text{sq}} = \frac{a^2}{1 - r^2} = 3 \quad \text{--- (2)}$$

Substitute $a^2 = 9(1-r)^2$ from (1) into (2): $$\frac{9(1 - r)^2}{(1 - r)(1 + r)} = 3 \implies \frac{9(1 - r)}{1 + r} = 3$$

$$3(1 - r) = 1 + r \implies 3 - 3r = 1 + r \implies 4r = 2 \implies r = \frac{1}{2}$$

From (1): $$a = 3\left(1 - \frac{1}{2}\right) = \frac{3}{2}$$

Therefore, $a = \frac{3}{2}$ and $r = \frac{1}{2}$. Correct option is **(c)**.

For any two positive real numbers $a$ and $b$:

$$A - G = \frac{a + b}{2} - \sqrt{ab} = \frac{a + b - 2\sqrt{ab}}{2} = \frac{(\sqrt{a} - \sqrt{b})^2}{2} \ge 0$$

Since the square of any real number is non-negative, $A - G \ge 0 \implies A \ge G$.

Specifically, for two **unequal** positive numbers ($a \ne b$), $(\sqrt{a} - \sqrt{b})^2 > 0$, hence $A > G$. (As per textbook key: **(d)**).

Therefore, correct option is **(d)**.

Present Value $P = 9000$, Rate of appreciation $r = 10\% = 0.10$, Time $n = 3$ years.

Using compound appreciation formula: $$V = P(1 + r)^n = 9000 \times (1 + 0.10)^3 = 9000 \times (1.1)^3$$

$$V = 9000 \times 1.331 = 11979$$

Therefore, its worth 3 years from now is ₹ 11,979. Correct option is **(c)**.

Principal $P = 5000$, Annual Rate $R = 6\%$, Time $T = 5$ years.

Compounded semi-annually: rate per period $i = \frac{6\%}{2} = 3\% = 0.03$, total conversion periods $n = 5 \times 2 = 10$.

$$A = P(1 + i)^n = 5000(1 + 0.03)^{10} = 5000(1.03)^{10}$$

Therefore, the value is $5000(1.03)^{10}$. Correct option is **(d)**.

1. Since $a, b, c$ are in A.P. $\implies 2b = a + c \implies b = \frac{a+c}{2}$ (b is the A.M. of $a$ and $c$).

2. Since $a, b, c$ are in G.P. $\implies b^2 = ac \implies b = \sqrt{ac}$ (b is the G.M. of $a$ and $c$).

Since $\text{A.M.} = \text{G.M.} \implies \frac{a+c}{2} = \sqrt{ac} \implies (\sqrt{a} - \sqrt{c})^2 = 0 \implies a = c$.

Substituting $a = c$ gives $b = \frac{a+a}{2} = a$.

Hence, $a = b = c$. Correct option is **(c)**.

Let $x$ be the number added to 3, 7, 15 so that $(3+x), (7+x), (15+x)$ form a G.P.

By property of G.P.: $$(7 + x)^2 = (3 + x)(15 + x)$$

$$49 + 14x + x^2 = 45 + 18x + x^2$$

$$49 + 14x = 45 + 18x \implies 4x = 4 \implies x = 1$$

The numbers become $4, 8, 16$ with common ratio $r = 2$.

Therefore, 1 should be added. Correct option is **(a)**.

The geometric mean (G.M.) between two positive numbers $a$ and $b$ is: $$\text{G.M.} = \sqrt{ab}$$

$$\text{G.M.} = \sqrt{3 \times 12} = \sqrt{36} = 6$$

Therefore, the geometric mean is 6. Correct option is **(b)**.

**Step 1: Formula for ratio of sums**

Let the first A.P. have first term $a_1$, common diff $d_1$ and the second have $a_2, d_2$.

$$\frac{S_n}{S'_n} = \frac{\frac{n}{2}[2a_1 + (n-1)d_1]}{\frac{n}{2}[2a_2 + (n-1)d_2]} = \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{5n + 4}{9n + 6}$$

Dividing numerator and denominator by 2: $$\frac{a_1 + \left(\frac{n-1}{2}\right)d_1}{a_2 + \left(\frac{n-1}{2}\right)d_2} = \frac{5n + 4}{9n + 6}$$

**Step 2: Matching for the $18^{\text{th}}$ term ratio**

The $18^{\text{th}}$ terms ratio is $\frac{T_{18}}{T'_{18}} = \frac{a_1 + 17d_1}{a_2 + 17d_2}$.

Equating coefficients: $\frac{n-1}{2} = 17 \implies n - 1 = 34 \implies n = 35$.

**Step 3: Calculating the ratio**

$$\frac{T_{18}}{T'_{18}} = \frac{5(35) + 4}{9(35) + 6} = \frac{175 + 4}{315 + 6} = \frac{179}{321}$$

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

**Step 1: Set up equations**

Let the G.P. be $a, ar, ar^2, \dots$.

1. Given sum of first two terms: $$a + ar = 36 \implies a(1 + r) = 36 \quad \text{--- (1)}$$

2. Given product of $1^{\text{st}}$ and $3^{\text{rd}}$ term is 9 times $2^{\text{nd}}$ term: $$a \cdot (ar^2) = 9(ar) \implies a^2 r^2 = 9ar \implies ar = 9 \implies a = \frac{9}{r} \quad \text{--- (2)}$$

**Step 2: Solve for $r$ and $a$**

Substitute $a = \frac{9}{r}$ in (1): $$\frac{9}{r}(1 + r) = 36 \implies \frac{1+r}{r} = 4 \implies 1 + r = 4r \implies 3r = 1 \implies r = \frac{1}{3}$$

Then $$a = \frac{9}{1/3} = 27$$.

**Step 3: Calculate sum of first 8 terms ($S_8$)**

$$S_8 = \frac{a(1 - r^8)}{1 - r} = \frac{27\left(1 - \left(\frac{1}{3}\right)^8\right)}{1 - \frac{1}{3}} = \frac{27\left(1 - \frac{1}{6561}\right)}{\frac{2}{3}} = 27 \times \frac{3}{2} \times \frac{6560}{6561} = \frac{81}{2} \times \frac{6560}{6561} = \frac{6560}{2 \times 81} = \frac{3280}{81}$$

**Answer**: The sum of the first 8 terms is **$\frac{3280}{81}$** (or $40\frac{40}{81}$).

Let $S_n = 7 + 77 + 777 + 7777 + \dots \text{ up to } n \text{ terms}$.

Factor out 7: $$S_n = 7(1 + 11 + 111 + 1111 + \dots)$$

Multiply and divide by 9: $$S_n = \frac{7}{9}(9 + 99 + 999 + 9999 + \dots)$$

Express each term as $(10^k - 1)$: $$S_n = \frac{7}{9}\left[(10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1)\right]$$

Regroup G.P. and constants: $$S_n = \frac{7}{9}\left[(10 + 10^2 + 10^3 + \dots + 10^n) - n\right]$$

Using G.P. sum $\frac{10(10^n - 1)}{10 - 1} = \frac{10(10^n - 1)}{9}$: $$S_n = \frac{7}{9}\left[\frac{10(10^n - 1)}{9} - n\right] = \frac{70}{81}(10^n - 1) - \frac{7n}{9}$$

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

**Step 1: Assume symmetrical terms**

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

Product $= \left(\frac{a}{r}\right) \cdot a \cdot (ar) = a^3 = 1 \implies a = 1$.

**Step 2: Sum of terms**

$$\frac{1}{r} + 1 + r = \frac{39}{10} \implies \frac{1 + r + r^2}{r} = \frac{39}{10}$$

$$10(1 + r + r^2) = 39r \implies 10r^2 - 29r + 10 = 0$$

Factor the quadratic equation: $$10r^2 - 25r - 4r + 10 = 0 \implies 5r(2r - 5) - 2(2r - 5) = 0 \implies (2r - 5)(5r - 2) = 0$$

$$r = \frac{5}{2} \quad \text{or} \quad r = \frac{2}{5}$$

**Step 3: Finding the terms**

• If $r = \frac{5}{2}$, the terms are: $\frac{1}{5/2}, 1, 1 \times \frac{5}{2} \implies \frac{2}{5}, 1, \frac{5}{2}$.

• If $r = \frac{2}{5}$, the terms are: $\frac{5}{2}, 1, \frac{2}{5}$.

**Answer**: Common ratio is $\frac{5}{2}$ or $\frac{2}{5}$; the terms are **$\frac{2}{5}, 1, \frac{5}{2}$**.

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

1. Given $T_3 - T_1 = 9$: $$ar^2 - a = 9 \implies a(r^2 - 1) = 9 \quad \text{--- (1)}$$

2. Given $T_2 - T_4 = 18$: $$ar - ar^3 = 18 \implies ar(1 - r^2) = 18 \implies -ar(r^2 - 1) = 18 \quad \text{--- (2)}$$

**Dividing (2) by (1):** $$\frac{-ar(r^2 - 1)}{a(r^2 - 1)} = \frac{18}{9} \implies -r = 2 \implies r = -2$$

**Finding $a$:** From (1): $$a((-2)^2 - 1) = 9 \implies a(4 - 1) = 9 \implies 3a = 9 \implies a = 3$$

**The 4 numbers are:**

• $T_1 = 3$

• $T_2 = 3(-2) = -6$

• $T_3 = 3(-2)^2 = 12$

• $T_4 = 3(-2)^3 = -24$

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

**Step 1: Given relation**

$$\text{A.M.} = 2 \times \text{G.M.} \implies \frac{a + b}{2} = 2\sqrt{ab} \implies \frac{a + b}{2\sqrt{ab}} = \frac{2}{1}$$

**Step 2: Applying Componendo and Dividendo**

$$\frac{(a + b) + 2\sqrt{ab}}{(a + b) - 2\sqrt{ab}} = \frac{2 + 1}{2 - 1} = \frac{3}{1}$$

$$\frac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = \frac{3}{1}$$

Taking square root on both sides ($a > b > 0$): $$\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \frac{\sqrt{3}}{1}$$

**Step 3: Applying Componendo and Dividendo again**

$$\frac{(\sqrt{a} + \sqrt{b}) + (\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b}) - (\sqrt{a} - \sqrt{b})} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$$

$$\frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \implies \frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$$

**Step 4: Squaring both sides**

$$\frac{a}{b} = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3} - 1)^2} = \frac{3 + 1 + 2\sqrt{3}}{3 + 1 - 2\sqrt{3}} = \frac{4 + 2\sqrt{3}}{4 - 2\sqrt{3}} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}}$$

Hence proved that $a : b = (2 + \sqrt{3}) : (2 - \sqrt{3})$.

Let the G.P. terms be $a, ar, ar^2, \dots, ar^{n-1}$.

1. **Sum ($S$):** $$S = \frac{a(r^n - 1)}{r - 1}$$

2. **Product ($P$):** $$P = a \cdot ar \cdot ar^2 \dots ar^{n-1} = a^n r^{1 + 2 + \dots + (n-1)} = a^n r^{\frac{n(n-1)}{2}} \implies P^2 = a^{2n} r^{n(n-1)}$$

3. **Sum of Reciprocals ($R$):** $$R = \frac{1}{a} + \frac{1}{ar} + \dots + \frac{1}{ar^{n-1}} = \frac{1}{a} \left[\frac{1 - (1/r)^n}{1 - (1/r)}\right] = \frac{1}{a} \left[\frac{r^n - 1}{r^n} \cdot \frac{r}{r - 1}\right] = \frac{r^n - 1}{a r^{n-1}(r - 1)}$$

Notice that $$R = \frac{S}{a^2 r^{n-1}} \implies R^n = \frac{S^n}{a^{2n} r^{n(n-1)}}$$

**Computing LHS ($P^2 R^n$):**

$$P^2 R^n = \left(a^{2n} r^{n(n-1)}\right) \times \frac{S^n}{a^{2n} r^{n(n-1)}} = S^n = \text{RHS}$$

Hence proved that $P^2 R^n = S^n$.

Given: Principal $P = 5000$, Interest rate $i = 8\% = 0.08$, Time $n = 10$ years.

Compounded annually formula: $$A = P(1 + i)^n = 5000(1 + 0.08)^{10} = 5000(1.08)^{10}$$

Using $(1.08)^{10} \approx 2.158925$:

$$A = 5000 \times 2.158925 = ₹ 10794.62$$

**Answer**: The amount will be **$5000(1.08)^{10}$** (approx **₹ 10,794.62**).

Let the G.P. be $T_1, T_2, \dots, T_n$ where $T_1 = a$ and $T_n = b = ar^{n-1}$.

The product of $n$ terms is: $$P = a \cdot (ar) \cdot (ar^2) \dots (ar^{n-1}) = a^n r^{1 + 2 + \dots + (n-1)} = a^n r^{\frac{n(n-1)}{2}}$$

Squaring both sides: $$P^2 = a^{2n} r^{n(n-1)} = \left(a^2 r^{n-1}\right)^n = \left(a \cdot ar^{n-1}\right)^n$$

Since $b = ar^{n-1}$, we have: $$P^2 = (a \cdot b)^n = (ab)^n$$

Hence proved that $P^2 = (ab)^n$.

Initial population $P_0 = 1,000,000 = 10^6$.

Doubling interval $= 20\text{ minutes}$. Total duration $= 3\text{ hours} = 180\text{ minutes}$.

Number of doubling cycles $n = \frac{180}{20} = 9$.

The population forms a G.P. with ratio $r = 2$: $$P = P_0 \cdot 2^n = 10^6 \times 2^9$$

Since $2^9 = 512$: $$P = 10^6 \times 512 = 512,000,000 = 512 \text{ million}$$

**Answer**: After 3 hours, there will be **512 million bacteria** ($5.12 \times 10^8$).

**Step 1: Determine perimeters of successive triangles**

• Triangle 1 side $s_1 = 24\text{ cm} \implies$ Perimeter $P_1 = 3 \times 24 = 72\text{ cm}$.

• By the midpoint theorem, the side of the next inscribed triangle is half: $s_2 = 12\text{ cm} \implies P_2 = 3 \times 12 = 36\text{ cm}$.

• Successive perimeters are halved each time: $P_3 = 18\text{ cm}, P_4 = 9\text{ cm}, \dots$

**Step 2: Infinite G.P. Sum**

First term $a = 72$, common ratio $r = \frac{1}{2} < 1$.

$$\text{Sum of all perimeters } S_\infty = \frac{a}{1 - r} = \frac{72}{1 - 1/2} = \frac{72}{1/2} = 144\text{ cm}$$

**Answer**: The sum of perimeters of all the triangles is **144 cm**.

**Step 1: Modeling as an Arithmetic Progression (A.P.)**

Day 1: $a = 1000$.

Daily decline $d = -50$.

The number of patients on day $n$ is: $$T_n = a + (n - 1)d$$

**Step 2: Finding when $T_n = 0$**

$$1000 + (n - 1)(-50) = 0 \implies 50(n - 1) = 1000 \implies n - 1 = 20 \implies n = 21$$

**Answer**: Yes, there will be a day with no patients, and it will be the **$21^{\text{st}}$ day** from the initial day.

**Step 1: Tracking downward and upward distances**

• Initial drop: $h_0 = 240\text{ cm}$ (downward only).

• $1^{\text{st}}$ rebound: rises $\frac{4}{5}(240) = 192\text{ cm}$ and falls $192\text{ cm}$ $\implies 2 \times 192$.

• $2^{\text{nd}}$ rebound: rises $\left(\frac{4}{5}\right)^2(240)$ and falls the same $\implies 2 \times 240 \times (4/5)^2$.

**Step 2: Total Distance Formula**

$$\text{Total Distance } D = h_0 + 2\left[h_0 r + h_0 r^2 + h_0 r^3 + \dots\right] = h_0 + 2 h_0 \left(\frac{r}{1 - r}\right) = h_0 \left(\frac{1 + r}{1 - r}\right)$$

Substitute $h_0 = 240\text{ cm}$ and $r = \frac{4}{5}$:

$$D = 240 \times \left(\frac{1 + 4/5}{1 - 4/5}\right) = 240 \times \left(\frac{9/5}{1/5}\right) = 240 \times 9 = 2160\text{ cm}$$

**Answer**: The total distance traveled before coming to rest is **$2160\text{ cm}$** (or $21.6\text{ meters}$).

**Step 1: Total letters mailed up to the $8^{\text{th}}$ set**

• $1^{\text{st}}$ set: 5 letters

• $2^{\text{nd}}$ set: $5 \times 5 = 5^2 = 25$ letters

• ...

• $8^{\text{th}}$ set: $5^8$ letters.

Total letters mailed in the whole chain up to $8^{\text{th}}$ round is the sum of G.P. with $a = 5, r = 5, n = 8$:

$$S_8 = \frac{a(r^8 - 1)}{r - 1} = \frac{5(5^8 - 1)}{5 - 1} = \frac{5(390625 - 1)}{4} = \frac{5 \times 390624}{4} = 5 \times 97656 = 488,280 \text{ letters}$$

*(Note: If calculating for the $8^{\text{th}}$ set alone, $5^8 = 390,625$ letters).*

**Step 2: Total cost calculation**

• Total cumulative postage for all 8 rounds: $488280 \times ₹ 0.50 = ₹ 244,140$.

• Postage for the $8^{\text{th}}$ round specifically: $390625 \times ₹ 0.50 = ₹ 195,312.50$.

**Answer**: Cumulative postage is **₹ 2,44,140** (or **₹ 1,95,312.50** for the $8^{\text{th}}$ set alone).

**Step 1: Economic Multiplier Process**

• Initial expenditure: $a = 70\% \text{ of } 30000 = 0.70 \times 30000 = ₹ 21,000$.

• Second round spending: $70\% \text{ of } 21000 = 0.70 \times 21000 = ₹ 14,700$.

• Subsequent rounds form an infinite G.P. with common ratio $r = 0.70$.

**Step 2: Infinite G.P. Sum**

$$\text{Total Spending } S_\infty = \frac{a}{1 - r} = \frac{21000}{1 - 0.70} = \frac{21000}{0.30} = \frac{210000}{3} = ₹ 70,000$$

**Answer**: The total amount spent on consumer goods is **₹ 70,000**.

**Step 1: Rate of depreciation**

Present worth $P = ₹ 51,000$.

Depreciation fraction $= \frac{1}{5} = 20\%$. Remaining value factor $r = 1 - \frac{1}{5} = \frac{4}{5} = 0.80$.

Time $n = 3$ years.

**Step 2: Future Worth Calculation**

$$V = P \left(1 - \frac{1}{5}\right)^3 = 51000 \times \left(\frac{4}{5}\right)^3 = 51000 \times \frac{64}{125}$$

$$V = 51000 \times 0.512 = ₹ 26,112$$

**Answer**: The machine will be worth **₹ 26,112** three years from now.

**Part (i): Terrace Sequence & Common Difference**

• Terrace 1 length $a = 20\text{ m}$.

• Each subsequent terrace is $1.5\text{ m}$ shorter $\implies d = -1.5\text{ m}$.

• Sequence: **$20, 18.5, 17, 15.5, \dots$** with **$d = -1.5\text{ m}$**.

**Part (ii): Length of topmost (12th) terrace**

• $$T_{12} = a + (12 - 1)d = 20 + 11(-1.5) = 20 - 16.5 = 3.5\text{ m}$$

• **Answer**: The length of the topmost terrace is **$3.5\text{ meters}$**.

**Part (iii): Total cost of solar panelling for 12 terraces**

• Total length of 12 terraces: $$S_{12} = \frac{12}{2}[a + l] = 6[20 + 3.5] = 6 \times 23.5 = 141\text{ m}$$

• Total Cost at ₹ 500 per meter: $$141 \times 500 = ₹ 70,500$$

• **Answer**: Total cost is **₹ 70,500**.

**Part (iv): Number of terraces for total length of 118 meters**

• $$S_n = \frac{n}{2}[2(20) + (n-1)(-1.5)] = 118$$

• $$\frac{n}{2}[40 - 1.5n + 1.5] = 118 \implies n(41.5 - 1.5n) = 236$$

• $$1.5n^2 - 41.5n + 236 = 0 \implies 3n^2 - 83n + 472 = 0$$

• Solving quadratic: $$n = \frac{83 \pm \sqrt{83^2 - 4(3)(472)}}{6} = \frac{83 \pm \sqrt{6889 - 5664}}{6} = \frac{83 \pm \sqrt{1225}}{6} = \frac{83 \pm 35}{6}$$

• $$n = \frac{83 - 35}{6} = \frac{48}{6} = 8$$ (or $n = 19.67$ which is discarded).

• **Answer**: Exactly **8 terraces** can be constructed.

**Part (i): G.P. for first three days and common ratio**

• Day 1: $a = 10$.

• Day 2: $10 \times 3 = 30$.

• Day 3: $30 \times 3 = 90$.

• Sequence: **$10, 30, 90$** with common ratio **$r = 3$**.

**Part (ii): Daily downloads on the 5th day**

• $$T_5 = a r^{5-1} = 10 \times 3^4 = 10 \times 81 = 810$$

• **Answer**: **810 downloads** will happen specifically on Day 5.

**Part (iii): Cumulative downloads by end of 6th day**

• $$S_6 = \frac{a(r^6 - 1)}{r - 1} = \frac{10(3^6 - 1)}{3 - 1} = \frac{10(729 - 1)}{2} = 5 \times 728 = 3640$$

• Since $3640 \ge 3500$, **Yes, they will successfully achieve the target** (by 140 downloads).

**Part (iv): Day when daily downloads cross 5,000**

• We check $T_n = 10 \times 3^{n-1} > 5000 \implies 3^{n-1} > 500$.

• For $n = 6$: $3^{6-1} = 3^5 = 243 < 500$ ($T_6 = 2430$).

• For $n = 7$: $3^{7-1} = 3^6 = 729 > 500$ ($T_7 = 7290 > 5000$).

• **Answer**: The daily downloads will cross 5,000 on the **$7^{\text{th}}$ day**.

**Part (i): Height reached after 1st rebound**

• $$h_1 = \frac{3}{4} \times 80 = 60\text{ meters}$$

**Part (ii): Height reached after 2nd rebound**

• $$h_2 = \frac{3}{4} \times 60 = 45\text{ meters}$$

**Part (iii): Total vertical distance for superball ($r = 3/4$)**

• Using infinite bouncing distance formula: $$D = h_0 \left(\frac{1 + r}{1 - r}\right) = 80 \times \left(\frac{1 + 3/4}{1 - 3/4}\right) = 80 \times \left(\frac{7/4}{1/4}\right) = 80 \times 7 = 560\text{ meters}$$

• **Answer**: The total distance traveled is **$560\text{ meters}$**.

**Part (iv): Total vertical distance for tennis ball ($r = 1/2$)**

• $$D = 80 \times \left(\frac{1 + 1/2}{1 - 1/2}\right) = 80 \times \left(\frac{3/2}{1/2}\right) = 80 \times 3 = 240\text{ meters}$$

• **Answer**: The tennis ball travels a total distance of **$240\text{ meters}$**.

**Part (i): Verification of G.P. for Stock A**

• Values: $100, 150, 225$.

• $\frac{150}{100} = 1.5$ and $\frac{225}{150} = 1.5$.

• Since consecutive ratios are equal, **Yes, it forms a G.P.** with common ratio **$r = 1.5$ (or $3/2$)**.

**Part (ii): AM-GM Inequality for distinct positive numbers**

• For distinct positive real numbers ($a \ne b > 0$), $$\text{A.M.} > \text{G.M.} \quad \left(\frac{a+b}{2} > \sqrt{ab}\right)$$

**Part (iii): Finding $a$ and $b$ for Stock B**

• $\frac{a + b}{2} = 25 \implies a + b = 50$

• $\sqrt{ab} = 20 \implies ab = 400$

• Quadratic equation: $x^2 - (a+b)x + ab = 0 \implies x^2 - 50x + 400 = 0$

• $(x - 40)(x - 10) = 0 \implies x = 40, 10$.

• **Answer**: The two values are **$a = 40, b = 10$** (or $a = 10, b = 40$).

**Part (iv): Inserting 2 Arithmetic Means between 10 and 40**

• Inserting $n = 2$ means between $a = 10$ and $b = 40$:

• $$d = \frac{40 - 10}{2 + 1} = \frac{30}{3} = 10$$

• $A_1 = 10 + 10 = 20$

• $A_2 = 10 + 2(10) = 30$

• **Answer**: The intermediate targets are **20 and 30** (forming AP: $10, 20, 30, 40$).

**Part (i): Value after 1 year**

• Initial Cost $P = ₹ 50,00,000$.

• Value factor $r = 1 - 0.20 = 0.80$.

• Value after 1 year: $$V_1 = 50,00,000 \times 0.80 = ₹ 40,00,000$$ (40 Lakhs).

**Part (ii): General expression after $n$ years**

• $$V_n = P(1 - d)^n = 5000000(0.8)^n$$ (or $5000000(1 - 0.20)^n$).

**Part (iii): Value at the end of 4th year**

• $$V_4 = 50,00,000 \times (0.8)^4 = 50,00,000 \times 0.4096 = ₹ 20,48,000$$ (20.48 Lakhs).

**Part (iv): Scrap Decision ($V_n < ₹ 15,00,000$)**

• Value after 5 years: $$V_5 = 2048000 \times 0.8 = ₹ 16,38,400$$ (Above ₹ 15,00,000).

• Value after 6 years: $$V_6 = 1638400 \times 0.8 = ₹ 13,10,720$$ (Drops below ₹ 15,00,000).

• **Answer**: They will sell the crane as scrap **after the $6^{\text{th}}$ year**.

• **Checking Assertion (A)**: If $x, y, z$ are in G.P. $\implies y^2 = xz$. Taking logarithm on both sides: $\log(y^2) = \log(xz) \implies 2\log y = \log x + \log z$. Thus, **Assertion (A) is true**.

• **Checking Reason (R)**: If $a, b, c$ are in G.P., then $b^2 = ac$. The condition $2b = a + c$ is for Arithmetic Progression (A.P.), not G.P. Thus, **Reason (R) is false**.

Therefore, Assertion (A) is true but Reason (R) is false. Correct option is **(c)**.

• **Checking Reason (R)**: An infinite geometric series converges and has a finite sum if and only if $|r| < 1$. Thus, **Reason (R) is true**.

• **Checking Assertion (A)**: For the given series, $a = 1$ and $r = \frac{3/2}{1} = 1.5 > 1$. Since $|r| > 1$, the series diverges to $+\infty$. The formula $\frac{a}{1-r} = \frac{1}{1 - 1.5} = -2$ cannot be applied because the series is non-convergent. Thus, **Assertion (A) is false**.

Therefore, Assertion (A) is false but Reason (R) is true. Correct option is **(d)**.

• **Checking Assertion (A)**: Given $a = 5, d = 2, n = 10$. Sum $S_{10} = \frac{10}{2}[2(5) + (10 - 1)2] = 5[10 + 18] = 5 \times 28 = 140$. Thus, **Assertion (A) is true**.

• **Checking Reason (R)**: The sum of the first $n$ odd natural numbers ($1 + 3 + 5 + \dots + (2n-1)$) is indeed $n^2$. Thus, **Reason (R) is true**.

• However, Reason (R) is a general formula for odd numbers starting from 1, whereas Assertion (A) is calculated using the general A.P. sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$. Hence, R is not the correct explanation of A.

Therefore, Both A and R are true but R is not the correct explanation of A. Correct option is **(b)**.

• **Checking Reason (R)**: If $a, b, c$ are in G.P. with common ratio $r$, then $b = ar$ and $c = ar^2$. Raising to any power $k$: $a^k, b^k = a^k r^k, c^k = a^k r^{2k}$, which forms a G.P. with ratio $r^k$. Thus, **Reason (R) is true**.

• **Checking Assertion (A)**: For $k = 2$, $a^2, b^2, c^2$ are in G.P. with common ratio $r^2$. Thus, **Assertion (A) is true**.

• Since Reason (R) gives the exact algebraic theorem directly implying Assertion (A), **R is the correct explanation of A**.

Therefore, correct option is **(a)**.

• **Checking Reason (R)**: In the sequence $5, 55, 555, \dots$, the ratio $\frac{55}{5} = 11$, whereas $\frac{555}{55} = \frac{111}{11} = 10.09 \ne 11$. Since the ratio is not constant, the sequence is not a G.P. Thus, **Reason (R) is true**.

• **Checking Assertion (A)**: Since the series is not directly a G.P., the formula $S_n = \frac{a(r^n-1)}{r-1}$ cannot be applied directly (it requires transformation to $9, 99, 999 \dots = 10^k - 1$). Thus, **Assertion (A) is true**.

• Reason (R) correctly explains why the G.P. sum formula cannot be directly applied to Assertion (A).

Therefore, Both A and R are true and R is the correct explanation of A. Correct option is **(a)**.