Arithmetic Progressions

Patterns in Nature and Daily Life

• Real-World Patterns: Many natural structures and real-world processes follow sequential numerical patterns:
  • Petals on a sunflower, holes in a honeycomb, grains on a maize cob, and spirals on pineapples or pine cones.
  • Job Salary Structure: A starting monthly salary of $\text{Rs } 8000$ with an annual increment of $\text{Rs } 500$ creates the yearly salary sequence: $\text{Rs } 8000, 8500, 9000, \dots$
  • Ladder Rungs: A ladder whose rung lengths decrease uniformly by $2\,\text{cm}$ from bottom to top, starting with a bottom rung of $45\,\text{cm}$, yields the rung length sequence (1st to 8th rung from bottom): $45\,\text{cm}, 43\,\text{cm}, 41\,\text{cm}, 39\,\text{cm}, 37\,\text{cm}, 35\,\text{cm}, 33\,\text{cm}, 31\,\text{cm}$.
  • Savings Scheme Growth: An investment of $\text{Rs } 8000$ where the amount becomes $\frac{5}{4}$ times itself every $3$ years yields maturity amounts after $3, 6, 9,$ and $12$ years of $\text{Rs } 10000, 12500, 15625, 19531.25$, respectively.
  • Unit Squares: Squares with side lengths $1, 2, 3, \dots$ units contain unit squares numbering $1^2, 2^2, 3^2, \dots$ (or $1, 4, 9, \dots$).
  • Money Box Savings: Depositing $\text{Rs } 100$ on a child's 1st birthday and increasing the deposit by $\text{Rs } 50$ each year creates the annual sequence: $\text{Rs } 100, 150, 200, 250, \dots$
  • Rabbit Population Growth (Fibonacci Pattern): Starting with a pair of young rabbits that produce a new pair every month starting from their second month (assuming no deaths), the total number of pairs at the start of months $1$ through $6$ is $1, 1, 2, 3, 5, 8$.

Fundamentals of Arithmetic Progressions (AP)

• Definition of an Arithmetic Progression (AP): An Arithmetic Progression is a sequence or list of numbers in which each term is obtained by adding a fixed number to the preceding term, except the first term.

• Term: Each individual number listed in an Arithmetic Progression is called a term.

• Common Difference ($d$): The constant fixed number added to each preceding term to obtain the next term is known as the common difference.

  • The common difference $d$ can be positive, negative, or zero.

• Mathematical Representation:

  • Let the terms of an AP be denoted by $a_1, a_2, a_3, \dots, a_n$.
  • The common difference $d$ satisfies:     $a_2 - a_1 = a_3 - a_2 = \dots = a_n - a_{n-1} = d$
  • In general, for any positive integer $k$:     $d = a_{k+1} - a_k$

• General Form of an AP:

  • An AP with first term $a$ and common difference $d$ is expressed generally as:     $a, a + d, a + 2d, a + 3d, \dots$

• Classification of APs:

  • Finite AP: An AP that contains a finite number of terms. A finite AP always possesses a distinct last term.
  • Example: Student heights in a morning assembly queue: $147\,\text{cm}, 148\,\text{cm}, 149\,\text{cm}, \dots, 157\,\text{cm}$.
  • Example: Daily minimum temperatures in January arranged in ascending order: $-3.1^\circ\text{C}, -3.0^\circ\text{C}, -2.9^\circ\text{C}, -2.8^\circ\text{C}, -2.7^\circ\text{C}, -2.6^\circ\text{C}, -2.5^\circ\text{C}$.
  • Example: Balance loan money after paying $5\%$ of a $\text{Rs } 1000$ loan monthly: $\text{Rs } 950, 900, 850, 800, \dots, 50$.
  • Example: Academic cash prizes for toppers of Classes I through XII: $\text{Rs } 200, 250, 300, 350, \dots, 750$.
  • Example: Cumulative savings of $\text{Rs } 50$ monthly over $10$ months: $\text{Rs } 50, 100, 150, 200, 250, 300, 350, 400, 450, 500$.
  • Infinite AP: An AP that has infinitely many terms. An infinite AP does not have a last term.
  • Example: $1, 2, 3, 4, \dots$
  • Example: $100, 70, 40, 10, \dots$
  • Example: $-3, -2, -1, 0, \dots$

• Requirements to Construct an AP: Knowing both the first term $a$ and the common difference $d$ is necessary and sufficient to uniquely construct the entire sequence:

  • If $a = 6$ and $d = 3$, the AP is $6, 9, 12, 15, \dots$
  • If $a = 6$ and $d = -3$, the AP is $6, 3, 0, -3, \dots$
  • If $a = -7$ and $d = -2$, the AP is $-7, -9, -11, -13, \dots$
  • If $a = 1.0$ and $d = 0.1$, the AP is $1.0, 1.1, 1.2, 1.3, \dots$
  • If $a = 0$ and $d = 1\frac{1}{2}$, the AP is $0, 1\frac{1}{2}, 3, 4\frac{1}{2}, 6, \dots$
  • If $a = 2$ and $d = 0$, the AP is $2, 2, 2, 2, \dots$

• Rule for Determining Common Difference ($d$): Always subtract the $k$th term from the $(k+1)$th term, even if the $(k+1)$th term is smaller than the $k$th term ($d = a_{k+1} - a_k$).

The General Term ($n$th Term) of an AP

• Derivation of the Formula:

  • Let $a_1 = a$ be the first term and $d$ be the common difference.
  • $a_2 = a + d = a + (2 - 1)d$
  • $a_3 = a_2 + d = (a + d) + d = a + 2d = a + (3 - 1)d$
  • $a_4 = a_3 + d = (a + 2d) + d = a + 3d = a + (4 - 1)d$
  • Continuing this process, the $n$th term $a_n$ is expressed as:     $a_n = a + (n - 1)d$

• Terminology:

  • $a_n$ is called the general term of the AP.
  • If an AP has $m$ terms in total, then $a_m$ represents the last term, often denoted by $l$.

Sum of First $n$ Terms of an AP

• Historical Context (Gauss's Approach):

  • Carl Friedrich Gauss evaluated the sum of integers from $1$ to $100$ by pairing terms:     $S = 1 + 2 + 3 + \dots + 99 + 100$$S = 100 + 99 + 98 + \dots + 2 + 1$$2S = (100 + 1) + (99 + 2) + \dots + (1 + 100) = 101 \times 100 = 10100$$S = \frac{10100}{2} = 5050$

• Derivation of the Sum Formula:

  • Let $S_n$ denote the sum of the first $n$ terms of an AP:     $S_n = a + (a + d) + (a + 2d) + \dots + [a + (n - 1)d]$
  • Writing terms in reverse order:     $S_n = [a + (n - 1)d] + [a + (n - 2)d] + \dots + (a + d) + a$
  • Summing term-wise:     $2S_n = n \times [2a + (n - 1)d]$$S_n = \frac{n}{2}[2a + (n - 1)d]$

• Alternative Forms of Sum Formula:

  • Rewriting $2a$ as $a + a$:     $S_n = \frac{n}{2}[a + a + (n - 1)d] = \frac{n}{2}(a + a_n)$
  • If $a_n = l$ (the last term of a finite AP):     $S_n = \frac{n}{2}(a + l)$
  • This form is used when the first and last terms are known, but the common difference $d$ is not explicitly given.

• Relation Between $a_n$ and $S_n$:

  • The $n$th term is equal to the difference between the sum of the first $n$ terms and the sum of the first $(n - 1)$ terms:     $a_n = S_n - S_{n-1}$

• Sum of First $n$ Positive Integers:

  • For the natural numbers $1, 2, 3, \dots, n$, where $a = 1$ and $l = n$:     $S_n = \frac{n(n + 1)}{2}$

Arithmetic Mean

• Definition: If three numbers $a$, $b$, and $c$ are in an Arithmetic Progression, then the middle term $b$ is defined as the arithmetic mean of $a$ and $c$
• Formula:$b - a = c - b \implies 2b = a + c \implies b = \frac{a + c}{2}$

Comprehensive Worked Examples

• Example 1: For the AP $\frac{3}{2}, \frac{1}{2}, -\frac{1}{2}, -\frac{3}{2}, \dots$, find $a$ and $d$

  • Solution: First term $a = \frac{3}{2}$. Common difference $d = \frac{1}{2} - \frac{3}{2} = -1$

• Example 2: Check which lists form an AP and calculate the next two terms:

  • (i) $4, 10, 16, 22, \dots$
  • $a_2 - a_1 = 6$, $a_3 - a_2 = 6$, $a_4 - a_3 = 6$. Constant $d = 6$. Forms an AP.
  • Next two terms: $22 + 6 = 28$ and $28 + 6 = 34$
  • (ii) $1, -1, -3, -5, \dots$
  • $a_2 - a_1 = -2$, $a_3 - a_2 = -2$, $a_4 - a_3 = -2$. Constant $d = -2$. Forms an AP.
  • Next two terms: $-5 + (-2) = -7$ and $-7 + (-2) = -9$
  • (iii) $-2, 2, -2, 2, -2, \dots$
  • $a_2 - a_1 = 2 - (-2) = 4$; $a_3 - a_2 = -2 - 2 = -4$. Since $a_2 - a_1 \neq a_3 - a_2$, it does not form an AP.
  • (iv) $1, 1, 1, 2, 2, 2, 3, 3, 3, \dots$
  • $a_2 - a_1 = 0$; $a_4 - a_3 = 2 - 1 = 1$. Differences are not constant, so it does not form an AP.

• Example 3: Find the 10th term of the AP $2, 7, 12, \dots$

  • Solution: $a = 2$, $d = 7 - 2 = 5$, $n = 10$
  • $a_{10} = a + (10 - 1)d = 2 + 9 \times 5 = 47$

• Example 4: Which term of the AP $21, 18, 15, \dots$ is $-81$? Is any term $0$?

  • Solution: $a = 21$, $d = 18 - 21 = -3$, $a_n = -81$
  • $-81 = 21 + (n - 1)(-3) \implies -81 = 24 - 3n \implies -105 = -3n \implies n = 35$
  • Thus, the 35th term is $-81$
  • For $a_n = 0$: $21 + (n - 1)(-3) = 0 \implies 3(n - 1) = 21 \implies n - 1 = 7 \implies n = 8$
  • Thus, the 8th term is $0$

• Example 5: Determine the AP whose 3rd term is $5$ and 7th term is $9$

  • Solution: $a_3 = a + 2d = 5$ (Eq. 1) and $a_7 = a + 6d = 9$ (Eq. 2)
  • Subtracting Eq. 1 from Eq. 2: $4d = 4 \implies d = 1$
  • Substituting $d = 1$ into Eq. 1: $a + 2(1) = 5 \implies a = 3$
  • Required AP is $3, 4, 5, 6, 7, \dots$

• Example 6: Check whether $301$ is a term of the list $5, 11, 17, 23, \dots$

  • Solution: $a_2 - a_1 = 6$, $a_3 - a_2 = 6$. Constant difference $d = 6$, first term $a = 5$
  • Let $a_n = 301 \implies 301 = 5 + (n - 1) \times 6 \implies 301 = 6n - 1 \implies 6n = 302 \implies n = \frac{302}{6} = \frac{151}{3}$
  • Since $n$ must be a positive integer and $\frac{151}{3}$ is not an integer, $301$ is not a term of the AP.

• Example 7: How many two-digit numbers are divisible by $3$?

  • Solution: List of two-digit numbers divisible by $3$: $12, 15, 18, \dots, 99$
  • Here $a = 12$, $d = 3$, $a_n = 99$
  • $99 = 12 + (n - 1) \times 3 \implies 87 = (n - 1) \times 3 \implies n - 1 = 29 \implies n = 30$
  • There are $30$ two-digit numbers divisible by $3$

• Example 8: Find the 11th term from the last term (towards the first term) of the AP $10, 7, 4, \dots, -62$

  • Solution Method 1: $a = 10$, $d = -3$, $l = -62$
  • $-62 = 10 + (n - 1)(-3) \implies -72 = (n - 1)(-3) \implies n - 1 = 24 \implies n = 25$
  • Total terms = $25$. The 11th term from the end corresponds to the $(25 - 11 + 1) = 15\text{th}$ term from the start.
  • $a_{15} = 10 + (15 - 1)(-3) = 10 + 14(-3) = 10 - 42 = -32$
  • Solution Method 2 (Reversing AP): Reverse the AP: $-62, -59, -56, \dots, 10$ where $a = -62$ and $d = 3$
  • $a_{11} = -62 + (11 - 1) \times 3 = -62 + 30 = -32$

• Example 9: A sum of $\text{Rs } 1000$ is invested at $8\%$ simple interest per year. Calculate interest at the end of each year, test if it forms an AP, and find interest after $30$ years.

  • Solution: $\text{Simple Interest} = \frac{P \times R \times T}{100}$
  • Year 1: $\frac{1000 \times 8 \times 1}{100} = \text{Rs } 80$
  • Year 2: $\frac{1000 \times 8 \times 2}{100} = \text{Rs } 160$
  • Year 3: $\frac{1000 \times 8 \times 3}{100} = \text{Rs } 240$
  • Sequence: $80, 160, 240, \dots$ with $a = 80$ and $d = 80$. Forms an AP.
  • Interest after $30$ years $= a_{30} = 80 + (30 - 1) \times 80 = 80 + 29 \times 80 = \text{Rs } 2400$

• Example 10: A flower bed has $23$ rose plants in the 1st row, $21$ in the 2nd, $19$ in the 3rd, and $5$ in the last row. Find total number of rows.

  • Solution: Sequence: $23, 21, 19, \dots, 5$
  • $a = 23$, $d = -2$, $a_n = 5$
  • $5 = 23 + (n - 1)(-2) \implies -18 = (n - 1)(-2) \implies n - 1 = 9 \implies n = 10$
  • There are $10$ rows in the flower bed.

• Example 11: Find the sum of the first $22$ terms of the AP $8, 3, -2, \dots$

  • Solution: $a = 8$, $d = 3 - 8 = -5$, $n = 22$
  • $S_{22} = \frac{22}{2}[2(8) + (22 - 1)(-5)] = 11[16 + 21(-5)] = 11[16 - 105] = 11(-89) = -979$

• Example 12: If the sum of the first $14$ terms of an AP is $1050$ and its first term is $10$, find the 20th term.

  • Solution: $S_{14} = 1050$, $n = 14$, $a = 10$
  • $1050 = \frac{14}{2}[2(10) + (14 - 1)d] = 7[20 + 13d] = 140 + 91d$
  • $910 = 91d \implies d = 10$
  • $a_{20} = 10 + (20 - 1) \times 10 = 10 + 190 = 200$

• Example 13: How many terms of the AP $24, 21, 18, \dots$ must be taken so that their sum is $78$?

  • Solution: $a = 24$, $d = -3$, $S_n = 78$
  • $78 = \frac{n}{2}[2(24) + (n - 1)(-3)] = \frac{n}{2}[48 - 3n + 3] = \frac{n}{2}[51 - 3n]$
  • $156 = 51n - 3n^2 \implies 3n^2 - 51n + 156 = 0 \implies n^2 - 17n + 52 = 0$
  • $(n - 4)(n - 13) = 0 \implies n = 4$ or $n = 13$
  • Remark: Both values are admissible. The sum of the 5th through 13th terms equals $0$ because positive and negative terms cancel each other out.

• Example 14:

  • (i) Find the sum of the first $1000$ positive integers.
  • $S_{1000} = \frac{1000}{2}(1 + 1000) = 500 \times 1001 = 500500$
  • (ii) Find the sum of the first $n$ positive integers.
  • $S_n = \frac{n(1 + n)}{2} = \frac{n(n + 1)}{2}$

• Example 15: Find the sum of the first $24$ terms of the list of numbers where $a_n = 3 + 2n$

  • Solution: $a_1 = 3 + 2(1) = 5$, $a_2 = 3 + 2(2) = 7$, $a_3 = 3 + 2(3) = 9$
  • List $5, 7, 9, 11, \dots$ forms an AP with $a = 5$, $d = 2$
  • $S_{24} = \frac{24}{2}[2(5) + (24 - 1)(2)] = 12[10 + 46] = 12 \times 56 = 672$

• Example 16: A TV manufacturer produced $600$ sets in Year 3 and $700$ sets in Year 7. Production grows uniformly by a fixed number each year.

  • Solution: $a_3 = a + 2d = 600$ and $a_7 = a + 6d = 700$
  • Subtracting equations yields $4d = 100 \implies d = 25$
  • Substituting $d = 25$ gives $a + 50 = 600 \implies a = 550$
  • (i) Production in 1st year = $550$
  • (ii) Production in 10th year $= a_{10} = 550 + 9(25) = 550 + 225 = 775$
  • (iii) Total production in first 7 years $= S_7 = \frac{7}{2}[2(550) + (7 - 1)(25)] = \frac{7}{2}[1100 + 150] = \frac{7}{2}(1250) = 4375$

Exercise Summaries and Practical Applications

• Exercise 5.1 Key Problems:

  • Taxi fare: $\text{Rs } 15$ for 1st km, $\text{Rs } 8$ per additional km. Sequence: $15, 23, 31, 39, \dots$ (Forms AP with $a = 15, d = 8$).
  • Air in cylinder: Vacuum pump removes $\frac{1}{4}$ remaining air each time. Remaining air sequence: $V, \frac{3}{4}V, \left(\frac{3}{4}\right)^2V, \dots$ (Does NOT form AP as ratios are constant, not differences).
  • Well digging cost: $\text{Rs } 150$ for 1st metre, rises by $\text{Rs } 50$ each subsequent metre. Sequence: $150, 200, 250, 300, \dots$ (Forms AP with $a = 150, d = 50$).
  • Compound Interest: $\text{Rs } 10000$ deposited at $8\%$ compound interest per annum. Balance sequence: $10000\left(1 + \frac{8}{100}\right)^1, 10000\left(1 + \frac{8}{100}\right)^2, \dots$ (Does NOT form AP).

• Exercise 5.2 Key Numerical Applications:

  • Find 31st term of AP with $a_{11} = 38$ and $a_{16} = 73$:
  • $a + 10d = 38$, $a + 15d = 73 \implies 5d = 35 \implies d = 7, a = -32$
  • $a_{31} = -32 + 30(7) = -32 + 210 = 178$
  • AP of $50$ terms, $a_3 = 12$, last term $a_{50} = 106$. Find 29th term:
  • $a + 2d = 12$, $a + 49d = 106 \implies 47d = 94 \implies d = 2, a = 8$
  • $a_{29} = 8 + 28(2) = 64$
  • Three-digit numbers divisible by $7$: First is $105$, last is $994$.
  • $994 = 105 + (n - 1)7 \implies 889 = (n - 1)7 \implies n - 1 = 127 \implies n = 128$
  • Multiples of $4$ between $10$ and $250$: First is $12$, last is $248$
  • $248 = 12 + (n - 1)4 \implies 236 = (n - 1)4 \implies n - 1 = 59 \implies n = 60$
  • Subba Rao Salary (1995 start at $\text{Rs } 5000$, annual increment $\text{Rs } 200$): Reaching $\text{Rs } 7000$
  • $7000 = 5000 + (n - 1)200 \implies 2000 = (n - 1)200 \implies n - 1 = 10 \implies n = 11$ (Year 2005)
  • Ramkali Savings: Saves $\text{Rs } 5$ in week 1, increases by $\text{Rs } 1.75$ weekly. Reaching $\text{Rs } 20.75$
  • $20.75 = 5 + (n - 1)1.75 \implies 15.75 = (n - 1)1.75 \implies n - 1 = 9 \implies n = 10$

• Exercise 5.3 Key Numerical Applications:

  • Construction Delay Penalty: $\text{Rs } 200$ for day 1, $\text{Rs } 250$ for day 2, $\text{Rs } 300$ for day 3 ($d = 50$). Penalty for $30$ days delay:
  • $S_{30} = \frac{30}{2}[2(200) + (30 - 1)50] = 15[400 + 1450] = 15 \times 1850 = \text{Rs } 27750$
  • Cash Prizes: Total $\text{Rs } 700$ for $7$ prizes, each $\text{Rs } 20$ less than preceding.
  • $S_7 = 700, n = 7, d = -20$
  • $700 = \frac{7}{2}[2a + 6(-20)] \implies 200 = 2a - 120 \implies 2a = 320 \implies a = 160$
  • Prize values: $\text{Rs } 160, 140, 120, 100, 80, 60, 40$
  • School Tree Planting: Classes I to XII, $3$ sections per class. Class $k$ plants $k$ trees per section ($3k$ trees total per class).
  • AP of trees planted per class level: $3, 6, 9, \dots, 36$
  • Total trees $S_{12} = \frac{12}{2}(3 + 36) = 6 \times 39 = 234$
  • Spiral Semicircles Length: Radii $0.5\,\text{cm}, 1.0\,\text{cm}, 1.5\,\text{cm}, 2.0\,\text{cm}, \dots$ for $13$ consecutive semicircles ($\pi = \frac{22}{7}$).
  • Perimeter of semicircle $l_k = \pi r_k$
  • Total length $S_{13} = \pi (0.5 + 1.0 + 1.5 + \dots \text{ to } 13 \text{ terms}) = \pi \times \frac{13}{2}[2(0.5) + 12(0.5)] = \pi \times \frac{13}{2}[1 + 6] = \frac{22}{7} \times \frac{13}{2} \times 7 = 143\,\text{cm}$
  • Log Stacking: $200$ total logs. Bottom row $20$, next $19$, next $18$.
  • $S_n = 200$, $a = 20$, $d = -1$
  • $200 = \frac{n}{2}[40 + (n - 1)(-1)] \implies 400 = n(41 - n) \implies n^2 - 41n + 400 = 0$
  • $(n - 16)(n - 25) = 0$
  • If $n = 25$, top row logs $a_{25} = 20 + 24(-1) = -4$ (Impossible).
  • Thus $n = 16$ rows. Top row logs $a_{16} = 20 + 15(-1) = 5$
  • Potato Race: Bucket at starting point, 1st potato $5\,\text{m}$ away, subsequent $9$ potatoes $3\,\text{m}$ apart. Total $10$ potatoes.
  • Distance for $k$th potato $= 2 \times [5 + (k - 1)3]$
  • Distance sequence: $2(5) = 10\,\text{m}$, $2(8) = 16\,\text{m}$, $2(11) = 22\,\text{m}, \dots$
  • AP with $a = 10$, $d = 6$, $n = 10$
  • Total distance $S_{10} = \frac{10}{2}[2(10) + 9(6)] = 5[20 + 54] = 5 \times 74 = 370\,\text{m}$

Advanced / Optional Exercise Problems

• First Negative Term of an AP:

  • AP: $121, 117, 113, \dots$
  • $a = 121$, $d = -4$
  • Set $a_n < 0 \implies 121 + (n - 1)(-4) < 0 \implies 121 - 4n + 4 < 0 \implies 125 < 4n \implies n > \frac{125}{4} = 31.25$
  • Smallest integer $n = 32$. The 32nd term is the first negative term.

• Term Relations and Sums:

  • Sum of 3rd and 7th terms is $6$, their product is $8$. Find sum of first 16 terms ($S_{16}$).
  • $a_3 + a_7 = (a + 2d) + (a + 6d) = 2a + 8d = 6 \implies a + 4d = 3 \implies a = 3 - 4d$
  • $a_3 \times a_7 = (a + 2d)(a + 6d) = 8$
  • Substitute $a = 3 - 4d$: $(3 - 2d)(3 + 2d) = 8 \implies 9 - 4d^2 = 8 \implies 4d^2 = 1 \implies d = \pm \frac{1}{2}$
  • Case 1: $d = \frac{1}{2} \implies a = 1$
    • $S_{16} = \frac{16}{2}\left[2(1) + 15\left(\frac{1}{2}\right)\right] = 8\left[2 + \frac{15}{2}\right] = 8 \times \frac{19}{2} = 76$
  • Case 2: $d = -\frac{1}{2} \implies a = 5$
    • $S_{16} = \frac{16}{2}\left[2(5) + 15\left(-\frac{1}{2}\right)\right] = 8\left[10 - \frac{15}{2}\right] = 8 \times \frac{5}{2} = 20$

• Ladder Wood Requirement:

  • Rungs decrease from $45\,\text{cm}$ at bottom to $25\,\text{cm}$ at top. Distance between top and bottom rungs $= 2\frac{1}{2}\,\text{m} = 250\,\text{cm}$. Rung spacing $= 25\,\text{cm}$.
  • Total number of rungs $n = \frac{250}{25} + 1 = 10 + 1 = 11$
  • First term $a = 45$, last term $l = 25$
  • Total length of wood required $= S_{11} = \frac{11}{2}(45 + 25) = \frac{11}{2}(70) = 385\,\text{cm}$

• House Numbering Equation:

  • Row of houses numbered consecutively $1$ to $49$. House numbered $x$ exists such that sum of house numbers preceding $x$ equals sum of house numbers following $x$.
  • $S_{x-1} = S_{49} - S_x$
  • (\frac{(x - 1)x}{2} = \frac{49 \times 50}{2} - \frac{x(x + 1)}{2})
  • (\frac{x^2 - x}{2} + \frac{x^2 + x}{2} = \frac{2450}{2} \implies \frac{2x^2}{2} = 1225 \implies x^2 = 1225 \implies x = 35)

• Football Terrace Concrete Volume:

  • Terrace has $15$ steps, each $50\,\text{m}$ long. Rise $=\frac{1}{4}\,\text{m}$, tread $=\frac{1}{2}\,\text{m}$.
  • Concrete volume for $k$th step $= \text{Length} \times \text{Tread} \times (k \times \text{Rise}) = 50 \times \frac{1}{2} \times \left(k \times \frac{1}{4}\right) = \frac{25}{4}k\,\text{m}^3$
  • Sequence of volumes: $\frac{25}{4}, 2 \times \frac{25}{4}, 3 \times \frac{25}{4}, \dots, 15 \times \frac{25}{4}$
  • Total Volume $= \sum_{k=1}^{15} \frac{25}{4}k = \frac{25}{4} \times \frac{15(16)}{2} = \frac{25}{4} \times 120 = 750\,\text{m}^3$