End-of-Chapter Exercises
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Overview
This page provides comprehensive Ch 8: Exploring Sequences and Progressions - End-of-Chapter Exercises. Practice high-level challenge questions on AP system systems, GP construction, consecutive number sums, bacteria growth rates, and custom Tribonacci/partial-sum sequences with step-by-step solutions.
Predicting What Comes Next: Advanced AP, GP & Sum Sequences
Q1: AP Term System Solver
Find the 31st term of an AP whose 11th term is $38$ and 16th term is $73$.
Let $a$ be the first term and $d$ be the common difference.
• $t_{11} = a + 10d = 38$ (Equation 1)
• $t_{16} = a + 15d = 73$ (Equation 2)
• $t_{11} = a + 10d = 38$ (Equation 1)
• $t_{16} = a + 15d = 73$ (Equation 2)
Subtract Equation 1 from Equation 2:
$$(a + 15d) - (a + 10d) = 73 - 38$$
$$5d = 35 \implies d = 7$$
$$(a + 15d) - (a + 10d) = 73 - 38$$
$$5d = 35 \implies d = 7$$
Substitute $d = 7$ into Equation 1:
$$a + 10(7) = 38 \implies a + 70 = 38 \implies a = -32$$
$$a + 10(7) = 38 \implies a + 70 = 38 \implies a = -32$$
Find the 31st term ($t_{31}$):
$$t_{31} = a + 30d = -32 + 30(7) = -32 + 210 = \mathbf{178}$$
$$t_{31} = a + 30d = -32 + 30(7) = -32 + 210 = \mathbf{178}$$
31st term = 178
Q2: Reconstruct AP from Conditions
Determine the AP whose third term is $16$ and whose 7th term exceeds the 5th term by $12$.
Let the AP terms be represented by $t_n = a + (n-1)d$.
• $t_3 = a + 2d = 16$ (Equation 1)
• $t_7 = t_5 + 12 \implies (a + 6d) = (a + 4d) + 12$
• $t_3 = a + 2d = 16$ (Equation 1)
• $t_7 = t_5 + 12 \implies (a + 6d) = (a + 4d) + 12$
Simplify the second relation:
$$6d - 4d = 12 \implies 2d = 12 \implies d = 6$$
$$6d - 4d = 12 \implies 2d = 12 \implies d = 6$$
Substitute $d = 6$ into Equation 1:
$$a + 2(6) = 16 \implies a + 12 = 16 \implies a = 4$$
$$a + 2(6) = 16 \implies a + 12 = 16 \implies a = 4$$
The AP starts with $a = 4$ and increments by $d = 6$:
Sequence: **$4, 10, 16, 22, \dots$**
Sequence: **$4, 10, 16, 22, \dots$**
AP: $4, 10, 16, 22, \dots$
Q3: Three-Digit Numbers Divisible by 7
How many three-digit numbers are divisible by $7$?
Step 1: Identify the limits of 3-digit numbers divisible by 7:
• Smallest 3-digit number is $100$. $\frac{100}{7} \approx 14.28 \implies 7 \times 15 = 105$. So first term $a = 105$.
• Largest 3-digit number is $999$. $\frac{999}{7} \approx 142.71 \implies 7 \times 142 = 994$. So last term $t_n = 994$.
• Smallest 3-digit number is $100$. $\frac{100}{7} \approx 14.28 \implies 7 \times 15 = 105$. So first term $a = 105$.
• Largest 3-digit number is $999$. $\frac{999}{7} \approx 142.71 \implies 7 \times 142 = 994$. So last term $t_n = 994$.
Step 2: Calculate number of terms ($n$):
The sequence is an AP: $105, 112, 119, \dots, 994$ with common difference $d = 7$.
$$t_n = a + (n - 1)d$$
$$994 = 105 + (n - 1)7$$
$$889 = (n - 1)7 \implies n - 1 = \frac{889}{7} = 127 \implies n = \mathbf{128}$$
The sequence is an AP: $105, 112, 119, \dots, 994$ with common difference $d = 7$.
$$t_n = a + (n - 1)d$$
$$994 = 105 + (n - 1)7$$
$$889 = (n - 1)7 \implies n - 1 = \frac{889}{7} = 127 \implies n = \mathbf{128}$$
128 three-digit numbers are divisible by 7
Q4: Multiples of 4 between 10 and 250
How many multiples of $4$ lie between $10$ and $250$?
Step 1: Identify the limits:
• Smallest multiple of 4 strictly $> 10$ is $12$ ($a = 12$).
• Largest multiple of 4 strictly $< 250$ is $248$ ($t_n = 248$).
• Smallest multiple of 4 strictly $> 10$ is $12$ ($a = 12$).
• Largest multiple of 4 strictly $< 250$ is $248$ ($t_n = 248$).
Step 2: Calculate $n$ in AP: $12, 16, 20, \dots, 248$ ($d = 4$):
$$248 = 12 + (n - 1)4$$
$$236 = (n - 1)4 \implies n - 1 = \frac{236}{4} = 59 \implies n = \mathbf{60}$$
$$248 = 12 + (n - 1)4$$
$$236 = (n - 1)4 \implies n - 1 = \frac{236}{4} = 59 \implies n = \mathbf{60}$$
60 multiples of 4 lie between 10 and 250
Q5: Find GP from Conditions
Find a GP for which the sum of the first two terms is $-4$ and the fifth term is $4$ times the third term.
Let the GP terms be $a, ar, ar^2, \dots$.
• Sum of first 2 terms: $a + ar = -4 \implies a(1 + r) = -4$ (Equation 1)
• 5th term is 4 times 3rd term: $t_5 = 4t_3 \implies a r^4 = 4 (a r^2)$
• Sum of first 2 terms: $a + ar = -4 \implies a(1 + r) = -4$ (Equation 1)
• 5th term is 4 times 3rd term: $t_5 = 4t_3 \implies a r^4 = 4 (a r^2)$
Assuming $a \ne 0, r \ne 0$:
$$r^2 = 4 \implies r = \pm 2$$
$$r^2 = 4 \implies r = \pm 2$$
Case 1: If $r = 2$:
Substitute into Equation 1:
$$a(1 + 2) = -4 \implies 3a = -4 \implies a = -4/3$$
GP: **$-4/3, -8/3, -16/3, \dots$**
Substitute into Equation 1:
$$a(1 + 2) = -4 \implies 3a = -4 \implies a = -4/3$$
GP: **$-4/3, -8/3, -16/3, \dots$**
Case 2: If $r = -2$:
Substitute into Equation 1:
$$a(1 - 2) = -4 \implies -a = -4 \implies a = 4$$
GP: **$4, -8, 16, -32, \dots$**
Substitute into Equation 1:
$$a(1 - 2) = -4 \implies -a = -4 \implies a = 4$$
GP: **$4, -8, 16, -32, \dots$**
Two possible GPs: (i) $a = -4/3, r = 2$ or (ii) $a = 4, r = -2$
Q6: Express 100 as Sum of Consecutive Naturals
Find all possible ways of expressing $100$ as the sum of consecutive natural numbers.
Let the sum of $k$ consecutive natural numbers starting from $a$ be equal to $100$:
$$a + (a+1) + (a+2) + \dots + (a+k-1) = 100$$
Using AP sum formula:
$$\frac{k}{2}[2a + (k-1)] = 100 \implies k[2a + k - 1] = 200$$
Where $a \ge 1$ (natural number) and $k \ge 2$ (at least two numbers).
$$a + (a+1) + (a+2) + \dots + (a+k-1) = 100$$
Using AP sum formula:
$$\frac{k}{2}[2a + (k-1)] = 100 \implies k[2a + k - 1] = 200$$
Where $a \ge 1$ (natural number) and $k \ge 2$ (at least two numbers).
Thus, $k$ must be a factor of 200. Let's analyze the factors of 200 ($1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 200$).
Also, since $2a = \frac{200}{k} - k + 1$, $\frac{200}{k} - k + 1$ must be a positive even integer.
Also, since $2a = \frac{200}{k} - k + 1$, $\frac{200}{k} - k + 1$ must be a positive even integer.
Let's test possible values of $k$:
- If $k = 5$: $2a = \frac{200}{5} - 5 + 1 = 40 - 4 = 36 \implies a = 18$.
Sum: $18 + 19 + 20 + 21 + 22 = 100$ (Valid!) - If $k = 8$: $2a = \frac{200}{8} - 8 + 1 = 25 - 7 = 18 \implies a = 9$.
Sum: $9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100$ (Valid!) - If $k = 15$ (not factor of 200), etc. Other factors of 200 like 2, 4, 10, 20, 25, 40 give non-integer or non-positive values for $a$.
For example, if $k=25$: $2a = 8 - 25 + 1 = -16$ (invalid). If $k=10$: $2a = 20 - 10 + 1 = 11$ (not even).
Two ways: (i) 5 terms: $18+19+20+21+22$, (ii) 8 terms: $9+10+11+12+13+14+15+16$
Q7: Bacteria Doubling GP Model
The number of bacteria in a certain culture doubles every hour. If there were $30\text{ bacteria}$ originally present in the culture, how many will be present at the end of the 2nd hour, 4th hour and $n$-th hour?
Let original count be $N_0 = 30$.
Since the population doubles every hour, this forms a GP:
• Count at start ($t=0$): $30$
• End of 1st hour ($t=1$): $30 \times 2 = 60$
• End of 2nd hour ($t=2$): $30 \times 2^2 = 120$
Since the population doubles every hour, this forms a GP:
• Count at start ($t=0$): $30$
• End of 1st hour ($t=1$): $30 \times 2 = 60$
• End of 2nd hour ($t=2$): $30 \times 2^2 = 120$
Using the formula $N_t = 30 \times 2^t$ for $t$ hours:
• **End of 2nd hour ($t=2$):** $30 \times 2^2 = \mathbf{120\text{ bacteria}}$
• **End of 4th hour ($t=4$):** $30 \times 2^4 = 30 \times 16 = \mathbf{480\text{ bacteria}}$
• **End of $n$-th hour ($t=n$):** $\mathbf{30 \times 2^n\text{ bacteria}}$
• **End of 2nd hour ($t=2$):** $30 \times 2^2 = \mathbf{120\text{ bacteria}}$
• **End of 4th hour ($t=4$):** $30 \times 2^4 = 30 \times 16 = \mathbf{480\text{ bacteria}}$
• **End of $n$-th hour ($t=n$):** $\mathbf{30 \times 2^n\text{ bacteria}}$
2nd hour: 120; 4th hour: 480; n-th hour: $30 \times 2^n$
Q8: AP Dual-Sum Solver
The sum of the 4th and 8th terms of an AP is $24$ and the sum of the 6th and 10th terms is $44$. Find the first three terms of the AP.
Write equations for the terms:
• $t_4 + t_8 = 24 \implies (a+3d) + (a+7d) = 24 \implies 2a + 10d = 24 \implies a + 5d = 12$ (Equation 1)
• $t_6 + t_10 = 44 \implies (a+5d) + (a+9d) = 44 \implies 2a + 14d = 44 \implies a + 7d = 22$ (Equation 2)
• $t_4 + t_8 = 24 \implies (a+3d) + (a+7d) = 24 \implies 2a + 10d = 24 \implies a + 5d = 12$ (Equation 1)
• $t_6 + t_10 = 44 \implies (a+5d) + (a+9d) = 44 \implies 2a + 14d = 44 \implies a + 7d = 22$ (Equation 2)
Subtract Equation 1 from Equation 2:
$$(a + 7d) - (a + 5d) = 22 - 12$$
$$2d = 10 \implies d = 5$$
$$(a + 7d) - (a + 5d) = 22 - 12$$
$$2d = 10 \implies d = 5$$
Substitute $d = 5$ into Equation 1:
$$a + 5(5) = 12 \implies a + 25 = 12 \implies a = -13$$
$$a + 5(5) = 12 \implies a + 25 = 12 \implies a = -13$$
First three terms:
• $t_1 = a = -13$
• $t_2 = a + d = -13 + 5 = -8$
• $t_3 = a + 2d = -13 + 10 = -3$
• $t_1 = a = -13$
• $t_2 = a + d = -13 + 5 = -8$
• $t_3 = a + 2d = -13 + 10 = -3$
First three terms: $-13, -8, -3$
Q9: Natural Number Sum Threshold
Find the smallest value of $n$ such that the sum of the first $n$ natural numbers is greater than $1,000$.
Set up the inequality:
$$S_n = \frac{n(n+1)}{2} > 1000 \implies n(n+1) > 2000$$
$$S_n = \frac{n(n+1)}{2} > 1000 \implies n(n+1) > 2000$$
Find the approximate root of $n^2 \approx 2000$:
Since $\sqrt{2000} \approx 44.7$:
• Try $n = 44$: $44 \times 45 = 1980$ (which is $< 2000$, so $S_{44} = 990$)
• Try $n = 45$: $45 \times 46 = 2070$ (which is $> 2000$, so $S_{45} = 1035$)
Since $\sqrt{2000} \approx 44.7$:
• Try $n = 44$: $44 \times 45 = 1980$ (which is $< 2000$, so $S_{44} = 990$)
• Try $n = 45$: $45 \times 46 = 2070$ (which is $> 2000$, so $S_{45} = 1035$)
Thus, the smallest value is **$n = 45$**.
Smallest $n = 45$
Q10: Large GP Term Position
Which term of the GP: $2, 8, 32, \dots$ is $131072$? Write the explicit formula as well as the recursive formula for the $n$-th term.
Identify first term $a = 2$ and common ratio $r = 4$.
• **Explicit formula:** $t_n = 2 \cdot 4^{n-1} = 2 \cdot (2^2)^{n-1} = 2 \cdot 2^{2n-2} = 2^{2n-1}$
• **Recursive formula:** $t_1 = 2, \quad t_n = 4t_{n-1} \text{ for } n \ge 2$
• **Explicit formula:** $t_n = 2 \cdot 4^{n-1} = 2 \cdot (2^2)^{n-1} = 2 \cdot 2^{2n-2} = 2^{2n-1}$
• **Recursive formula:** $t_1 = 2, \quad t_n = 4t_{n-1} \text{ for } n \ge 2$
Set $t_n = 131072$ and solve for $n$:
$$2^{2n-1} = 131072$$
Since $131072 = 2^{17}$:
$$2n - 1 = 17 \implies 2n = 18 \implies n = 9$$
$$2^{2n-1} = 131072$$
Since $131072 = 2^{17}$:
$$2n - 1 = 17 \implies 2n = 18 \implies n = 9$$
131072 is the 9th term. Explicit: $t_n = 2 \cdot 4^{n-1}$
Q11: GP System of Product and Sum
The sum of the first three terms of a GP is $\frac{13}{12}$ and their product is $-1$. Find the common ratio and the terms.
Let the three terms of the GP be $\frac{a}{r}, a, ar$.
• Product: $\left(\frac{a}{r}\right) \times a \times (ar) = -1 \implies a^3 = -1 \implies a = -1$.
• Product: $\left(\frac{a}{r}\right) \times a \times (ar) = -1 \implies a^3 = -1 \implies a = -1$.
Substitute $a = -1$ into the sum relation:
$$\frac{-1}{r} - 1 - r = \frac{13}{12} \implies -\left(\frac{1}{r} + 1 + r\right) = \frac{13}{12}$$
$$\frac{r^2 + r + 1}{r} = -\frac{13}{12} \implies 12r^2 + 12r + 12 = -13r \implies 12r^2 + 25r + 12 = 0$$
$$\frac{-1}{r} - 1 - r = \frac{13}{12} \implies -\left(\frac{1}{r} + 1 + r\right) = \frac{13}{12}$$
$$\frac{r^2 + r + 1}{r} = -\frac{13}{12} \implies 12r^2 + 12r + 12 = -13r \implies 12r^2 + 25r + 12 = 0$$
Solve the quadratic equation using middle-term splitting:
$$12r^2 + 16r + 9r + 12 = 0$$
$$4r(3r + 4) + 3(3r + 4) = 0 \implies (4r + 3)(3r + 4) = 0$$
$$r = -3/4 \quad \text{or} \quad r = -4/3$$
$$12r^2 + 16r + 9r + 12 = 0$$
$$4r(3r + 4) + 3(3r + 4) = 0 \implies (4r + 3)(3r + 4) = 0$$
$$r = -3/4 \quad \text{or} \quad r = -4/3$$
If $r = -3/4$:
Terms: $\frac{-1}{-3/4}, -1, -1(-3/4) \implies \mathbf{4/3, -1, 3/4}$
If $r = -4/3$:
Terms: $\frac{-1}{-4/3}, -1, -1(-4/3) \implies \mathbf{3/4, -1, 4/3}$
Terms: $\frac{-1}{-3/4}, -1, -1(-3/4) \implies \mathbf{4/3, -1, 3/4}$
If $r = -4/3$:
Terms: $\frac{-1}{-4/3}, -1, -1(-4/3) \implies \mathbf{3/4, -1, 4/3}$
Common ratio $r = -3/4$ or $-4/3$; Terms are $4/3, -1, 3/4$
Q12: Prove Term Relations in GP
If the 4th, 10th and 16th terms of a GP are $x, y$ and $z$ respectively, prove that $x, y, z$ are in GP.
Let the GP have first term $a$ and common ratio $r$.
• $x = t_4 = a r^3$
• $y = t_{10} = a r^9$
• $z = t_{16} = a r^{15}$
• $x = t_4 = a r^3$
• $y = t_{10} = a r^9$
• $z = t_{16} = a r^{15}$
To prove $x, y, z$ are in GP, we must show that $y^2 = xz$:
$$y^2 = (a r^9)^2 = a^2 r^{18}$$
$$xz = (a r^3) \cdot (a r^{15}) = a^2 r^{3+15} = a^2 r^{18}$$
$$y^2 = (a r^9)^2 = a^2 r^{18}$$
$$xz = (a r^3) \cdot (a r^{15}) = a^2 r^{3+15} = a^2 r^{18}$$
Since $y^2 = xz$, the terms $x, y,$ and $z$ form a Geometric Progression.
Proved: $y^2 = xz$
Q13: GP Squares System
The sum of the first three terms of a geometric progression is $26$, and the sum of their squares is $364$. Find the terms of the GP.
Let the terms be $a, ar, ar^2$.
• $a(1 + r + r^2) = 26$ (Equation 1)
• $a^2(1 + r^2 + r^4) = 364$ (Equation 2)
• $a(1 + r + r^2) = 26$ (Equation 1)
• $a^2(1 + r^2 + r^4) = 364$ (Equation 2)
Divide Equation 2 by the square of Equation 1:
$$\frac{a^2(1 + r^2 + r^4)}{a^2(1 + r + r^2)^2} = \frac{364}{26^2} = \frac{364}{676} = \frac{7}{13}$$
Since $1 + r^2 + r^4 = (1 + r + r^2)(1 - r + r^2)$:
$$\frac{1 - r + r^2}{1 + r + r^2} = \frac{7}{13}$$
$$\frac{a^2(1 + r^2 + r^4)}{a^2(1 + r + r^2)^2} = \frac{364}{26^2} = \frac{364}{676} = \frac{7}{13}$$
Since $1 + r^2 + r^4 = (1 + r + r^2)(1 - r + r^2)$:
$$\frac{1 - r + r^2}{1 + r + r^2} = \frac{7}{13}$$
Cross-multiply and solve for $r$:
$$13(1 - r + r^2) = 7(1 + r + r^2)$$
$$13 - 13r + 13r^2 = 7 + 7r + 7r^2 \implies 6r^2 - 20r + 6 = 0 \implies 3r^2 - 10r + 3 = 0$$
Solving this quadratic yields $r = 3$ or $r = 1/3$.
$$13(1 - r + r^2) = 7(1 + r + r^2)$$
$$13 - 13r + 13r^2 = 7 + 7r + 7r^2 \implies 6r^2 - 20r + 6 = 0 \implies 3r^2 - 10r + 3 = 0$$
Solving this quadratic yields $r = 3$ or $r = 1/3$.
Substitute $r = 3$ into Equation 1:
$$a(1 + 3 + 9) = 26 \implies 13a = 26 \implies a = 2$$
The terms are **$2, 6, 18$**.
$$a(1 + 3 + 9) = 26 \implies 13a = 26 \implies a = 2$$
The terms are **$2, 6, 18$**.
Terms are $2, 6, 18$ (or $18, 6, 2$)
Q14: Cumulative Recursive Sequence Rule
Suppose $P_1 = 1, P_2 = 2$ and for $n > 2$, $P_n = P_1 + P_2 + \dots + P_{n-1} + 1$.
(i) Find the values of $P_1, P_2, \dots, P_8$.
(ii) Can you find a simpler recursive formula for $P_n$?
(iii) Can you give an explicit formula?
(i) Find the values of $P_1, P_2, \dots, P_8$.
(ii) Can you find a simpler recursive formula for $P_n$?
(iii) Can you give an explicit formula?
(i) Calculate first 8 terms:
• $P_1 = 1$
• $P_2 = 2$
• $P_3 = P_1 + P_2 + 1 = 1 + 2 + 1 = 4$
• $P_4 = (P_1 + P_2 + P_3) + 1 = (1 + 2 + 4) + 1 = 8$
• $P_5 = (1 + 2 + 4 + 8) + 1 = 16$
• $P_6 = 32$
• $P_7 = 64$
• $P_8 = 128$
• $P_1 = 1$
• $P_2 = 2$
• $P_3 = P_1 + P_2 + 1 = 1 + 2 + 1 = 4$
• $P_4 = (P_1 + P_2 + P_3) + 1 = (1 + 2 + 4) + 1 = 8$
• $P_5 = (1 + 2 + 4 + 8) + 1 = 16$
• $P_6 = 32$
• $P_7 = 64$
• $P_8 = 128$
(ii) Simpler Recursive Formula:
Observe that for $n \ge 3$, each term is exactly twice the previous term:
$$P_n = P_1 + \dots + P_{n-2} + P_{n-1} + 1$$
Since $P_{n-1} = P_1 + \dots + P_{n-2} + 1$:
$$P_n = P_{n-1} + P_{n-1} = 2P_{n-1}$$
So simpler rule: **$P_1 = 1, P_2 = 2, P_n = 2P_{n-1}$ for $n \ge 3$**.
Observe that for $n \ge 3$, each term is exactly twice the previous term:
$$P_n = P_1 + \dots + P_{n-2} + P_{n-1} + 1$$
Since $P_{n-1} = P_1 + \dots + P_{n-2} + 1$:
$$P_n = P_{n-1} + P_{n-1} = 2P_{n-1}$$
So simpler rule: **$P_1 = 1, P_2 = 2, P_n = 2P_{n-1}$ for $n \ge 3$**.
(iii) Explicit Formula:
Observe the sequence is: $1, 2, 4, 8, 16, 32, \dots$.
This is exactly powers of 2:
$$\mathbf{P_n = 2^{n-1} \text{ for } n \ge 1}$$
Observe the sequence is: $1, 2, 4, 8, 16, 32, \dots$.
This is exactly powers of 2:
$$\mathbf{P_n = 2^{n-1} \text{ for } n \ge 1}$$
Sequence: $1, 2, 4, 8, 16, 32, 64, 128$; Simplest recursive: $P_n = 2P_{n-1}$ ($n \ge 3$); Explicit: $P_n = 2^{n-1}$
Q15: Delayed Summation Sequence Recognition
Suppose $W_1 = 1, W_2 = 2$ and for $n > 2$, $W_n = W_1 + W_2 + \dots + W_{n-2} + 2$.
(i) Find the values of $W_1, W_2, \dots, W_8$.
(ii) Do you recognise this sequence?
(i) Find the values of $W_1, W_2, \dots, W_8$.
(ii) Do you recognise this sequence?
(i) Calculate first 8 terms:
• $W_1 = 1$
• $W_2 = 2$
• $W_3 = W_1 + 2 = 1 + 2 = \mathbf{3}$
• $W_4 = W_1 + W_2 + 2 = 1 + 2 + 2 = \mathbf{5}$
• $W_5 = W_1 + W_2 + W_3 + 2 = 1 + 2 + 3 + 2 = \mathbf{8}$
• $W_6 = W_1 + W_2 + W_3 + W_4 + 2 = 1 + 2 + 3 + 5 + 2 = \mathbf{13}$
• $W_7 = W_1 + W_2 + W_3 + W_4 + W_5 + 2 = 1 + 2 + 3 + 5 + 8 + 2 = \mathbf{21}$
• $W_8 = W_1 + W_2 + W_3 + W_4 + W_5 + W_6 + 2 = 1 + 2 + 3 + 5 + 8 + 13 + 2 = \mathbf{34}$
• $W_1 = 1$
• $W_2 = 2$
• $W_3 = W_1 + 2 = 1 + 2 = \mathbf{3}$
• $W_4 = W_1 + W_2 + 2 = 1 + 2 + 2 = \mathbf{5}$
• $W_5 = W_1 + W_2 + W_3 + 2 = 1 + 2 + 3 + 2 = \mathbf{8}$
• $W_6 = W_1 + W_2 + W_3 + W_4 + 2 = 1 + 2 + 3 + 5 + 2 = \mathbf{13}$
• $W_7 = W_1 + W_2 + W_3 + W_4 + W_5 + 2 = 1 + 2 + 3 + 5 + 8 + 2 = \mathbf{21}$
• $W_8 = W_1 + W_2 + W_3 + W_4 + W_5 + W_6 + 2 = 1 + 2 + 3 + 5 + 8 + 13 + 2 = \mathbf{34}$
(ii) Recognition:
The sequence of terms is **$1, 2, 3, 5, 8, 13, 21, 34, \dots$**.
This is the famous **Virahānka-Fibonacci sequence**! Each term for $n \ge 3$ is the sum of the preceding two terms: $W_n = W_{n-1} + W_{n-2}$.
The sequence of terms is **$1, 2, 3, 5, 8, 13, 21, 34, \dots$**.
This is the famous **Virahānka-Fibonacci sequence**! Each term for $n \ge 3$ is the sum of the preceding two terms: $W_n = W_{n-1} + W_{n-2}$.
Sequence: $1, 2, 3, 5, 8, 13, 21, 34$; This is the Virahānka-Fibonacci sequence