Textbook -Arithmetic Progressions Questions in English

(A) The positive integers that are divisible by $6$ are $6, 12, 18, 24, \ldots$
This sequence forms an Arithmetic Progression $(A.P.)$ where the first term $a = 6$ and the common difference $d = 6$.
We need to find the sum of the first $n = 40$ terms.
The formula for the sum of the first $n$ terms of an $A.P.$ is $S_n = \frac{n}{2}[2a + (n - 1)d]$.
Substituting the values $n = 40$,$a = 6$,and $d = 6$:
$S_{40} = \frac{40}{2}[2(6) + (40 - 1)6]$
$S_{40} = 20[12 + (39)(6)]$
$S_{40} = 20[12 + 234]$
$S_{40} = 20 \times 246$
$S_{40} = 4920$.

(B) The multiples of $8$ are $8, 16, 24, 32, \ldots$
These numbers form an Arithmetic Progression $(A.P.)$ where the first term $a = 8$ and the common difference $d = 8$.
We need to find the sum of the first $n = 15$ terms.
The formula for the sum of the first $n$ terms of an $A.P.$ is $S_n = \frac{n}{2}[2a + (n - 1)d]$.
Substituting the values: $S_{15} = \frac{15}{2}[2(8) + (15 - 1)8]$.
$S_{15} = \frac{15}{2}[16 + 14(8)]$.
$S_{15} = \frac{15}{2}[16 + 112]$.
$S_{15} = \frac{15}{2}(128)$.
$S_{15} = 15 \times 64 = 960$.

(C) The odd numbers between $0$ and $50$ are $1, 3, 5, 7, 9, \ldots, 49$.
These numbers form an Arithmetic Progression $(A.P.)$ where the first term $a = 1$,the common difference $d = 2$,and the last term $l = 49$.
Using the formula for the $n^{th}$ term: $l = a + (n - 1)d$.
$49 = 1 + (n - 1)2$
$48 = 2(n - 1)$
$n - 1 = 24$
$n = 25$.
Now,using the sum formula $S_n = \frac{n}{2}(a + l)$:
$S_{25} = \frac{25}{2}(1 + 49)$
$S_{25} = \frac{25}{2}(50)$
$S_{25} = 25 \times 25 = 625$.

(D) It can be observed that these penalties form an Arithmetic Progression $(A.P.)$ with the first term $a = 200$ and common difference $d = 50$.
The number of days is $n = 30$.
The total penalty to be paid is the sum of the first $30$ terms of the $A.P.$,given by the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$.
Substituting the values:
$S_{30} = \frac{30}{2}[2(200) + (30 - 1)50]$
$S_{30} = 15[400 + 29 \times 50]$
$S_{30} = 15[400 + 1450]$
$S_{30} = 15[1850]$
$S_{30} = 27750$
Therefore,the contractor has to pay ₹ $27750$ as a penalty.

(N/A) Let the value of the $1^{\text{st}}$ prize be $a$.
Since each prize is ₹ $20$ less than the preceding one,the values form an Arithmetic Progression $(A.P.)$ with common difference $d = -20$.
The number of prizes is $n = 7$ and the total sum is $S_7 = 700$.
The formula for the sum of $n$ terms of an $A.P.$ is $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values: $700 = \frac{7}{2} [2a + (7 - 1)(-20)]$.
$700 = \frac{7}{2} [2a + 6(-20)]$.
$100 = \frac{1}{2} [2a - 120]$.
$100 = a - 60$.
$a = 160$.
The values of the seven prizes are:
$1^{\text{st}}$ prize: ₹ $160$
$2^{\text{nd}}$ prize: ₹ $140$
$3^{\text{rd}}$ prize: ₹ $120$
$4^{\text{th}}$ prize: ₹ $100$
$5^{\text{th}}$ prize: ₹ $80$
$6^{\text{th}}$ prize: ₹ $60$
$7^{\text{th}}$ prize: ₹ $40$.

(B) It can be observed that the number of trees planted by the students of one section of each class forms an arithmetic progression $(AP)$.
The sequence is $1, 2, 3, 4, \dots, 12$.
Here,the first term $a = 1$ and the common difference $d = 2 - 1 = 1$.
The number of terms $n = 12$.
The sum of trees planted by one section of all classes is given by the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$.
$S_{12} = \frac{12}{2}[2(1) + (12 - 1)(1)]$
$S_{12} = 6[2 + 11]$
$S_{12} = 6 \times 13 = 78$.
Since there are $3$ sections for each class,the total number of trees planted by all students is $3 \times 78 = 234$.
Therefore,the total number of trees planted by the students is $234$.

(C) The length of a semicircle is given by $l = \pi r$.
For the given radii $r_1 = 0.5 \, cm, r_2 = 1.0 \, cm, r_3 = 1.5 \, cm, \ldots$,the lengths of the successive semicircles are:
$l_1 = \pi(0.5) = 0.5\pi \, cm$
$l_2 = \pi(1.0) = 1.0\pi \, cm$
$l_3 = \pi(1.5) = 1.5\pi \, cm$
These lengths form an Arithmetic Progression $(A.P.)$ where the first term $a = 0.5\pi$ and the common difference $d = 1.0\pi - 0.5\pi = 0.5\pi$.
We need to find the sum of $n = 13$ terms.
The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values:
$S_{13} = \frac{13}{2} [2(0.5\pi) + (13 - 1)(0.5\pi)]$
$S_{13} = \frac{13}{2} [1.0\pi + 12(0.5\pi)]$
$S_{13} = \frac{13}{2} [1.0\pi + 6.0\pi] = \frac{13}{2} [7\pi]$
$S_{13} = \frac{91\pi}{2}$
Using $\pi = \frac{22}{7}$:
$S_{13} = \frac{91}{2} \times \frac{22}{7} = 13 \times 11 = 143 \, cm$.
Thus,the total length of the spiral is $143 \, cm$.

(N/A) It can be observed that the numbers of logs in rows form an $A.P.$
$20, 19, 18, \dots$
For this $A.P.$
$a = 20$
$d = a_2 - a_1 = 19 - 20 = -1$
Let a total of $200$ logs be placed in $n$ rows.
$S_n = 200$
Using the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
$200 = \frac{n}{2}[2(20) + (n - 1)(-1)]$
$400 = n(40 - n + 1)$
$400 = n(41 - n)$
$400 = 41n - n^2$
$n^2 - 41n + 400 = 0$
$n^2 - 16n - 25n + 400 = 0$
$n(n - 16) - 25(n - 16) = 0$
$(n - 16)(n - 25) = 0$
So,$n = 16$ or $n = 25$.
Now,find the number of logs in the $n^{th}$ row using $a_n = a + (n - 1)d$:
For $n = 16$:
$a_{16} = 20 + (16 - 1)(-1) = 20 - 15 = 5$
For $n = 25$:
$a_{25} = 20 + (25 - 1)(-1) = 20 - 24 = -4$
Since the number of logs cannot be negative,$n = 25$ is rejected.
Therefore,the $200$ logs are placed in $16$ rows,and there are $5$ logs in the top row.

(A) The distances of the potatoes from the bucket are $5\, m, 8\, m, 11\, m, 14\, m, \dots$
This forms an Arithmetic Progression $(A.P.)$ where the first term $a = 5$ and the common difference $d = 3$.
For each potato,the competitor runs to the potato and back to the bucket,so the distance covered for each potato is twice its distance from the bucket.
The distances to be run for each of the $10$ potatoes are $2 \times 5, 2 \times 8, 2 \times 11, 2 \times 14, \dots$
This is a new $A.P.$ with first term $a' = 10$ and common difference $d' = 6$.
The total distance is the sum of the first $10$ terms of this $A.P.$
Using the sum formula $S_n = \frac{n}{2}[2a' + (n-1)d']$:
$S_{10} = \frac{10}{2}[2(10) + (10-1)6]$
$S_{10} = 5[20 + 9 \times 6]$
$S_{10} = 5[20 + 54]$
$S_{10} = 5[74] = 370$.
Thus,the total distance the competitor has to run is $370\, m$.

(B) Given $A.P.$ is $121, 117, 113, \ldots$
Here,the first term $a = 121$ and the common difference $d = 117 - 121 = -4$.
The $n^{\text{th}}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d$.
Substituting the values,we get $a_n = 121 + (n - 1)(-4) = 121 - 4n + 4 = 125 - 4n$.
We want to find the first negative term,so we set $a_n < 0$.
$125 - 4n < 0$
$125 < 4n$
$n > \frac{125}{4}$
$n > 31.25$
Since $n$ must be a natural number,the smallest integer greater than $31.25$ is $32$.
Therefore,the $32^{\text{nd}}$ term is the first negative term of this $A.P.$

(C) We know that the $n^{th}$ term of an $AP$ is $a_n = a + (n-1)d$.
Thus,$a_3 = a + 2d$ and $a_7 = a + 6d$.
Given $a_3 + a_7 = 6$,we have $(a + 2d) + (a + 6d) = 6$,which simplifies to $2a + 8d = 6$,or $a + 4d = 3$. So,$a = 3 - 4d$ $...(i)$.
Also,given $a_3 \times a_7 = 8$,we have $(a + 2d)(a + 6d) = 8$.
Substituting $a = 3 - 4d$ from $(i)$ into this equation:
$(3 - 4d + 2d)(3 - 4d + 6d) = 8$
$(3 - 2d)(3 + 2d) = 8$
$9 - 4d^2 = 8 \implies 4d^2 = 1 \implies d^2 = 1/4 \implies d = \pm 1/2$.
Case $1$: If $d = 1/2$,then $a = 3 - 4(1/2) = 3 - 2 = 1$.
The sum of the first $16$ terms is $S_{16} = \frac{16}{2}[2(1) + (16-1)(1/2)] = 8[2 + 7.5] = 8[9.5] = 76$.
Case $2$: If $d = -1/2$,then $a = 3 - 4(-1/2) = 3 + 2 = 5$.
The sum of the first $16$ terms is $S_{16} = \frac{16}{2}[2(5) + (16-1)(-1/2)] = 8[10 - 7.5] = 8[2.5] = 20$.
Since $20$ is one of the options,the correct answer is $20$.

(D) The rungs are placed at a distance of $25 \, cm$ from each other. The total distance between the top and bottom rung is $2 \frac{1}{2} \, m = 250 \, cm$.
The number of gaps between the rungs is $\frac{250}{25} = 10$.
Since there is a rung at both ends,the total number of rungs $n$ is $10 + 1 = 11$.
As the lengths of the rungs decrease uniformly,they form an Arithmetic Progression $(A.P.)$.
The first term $a = 45 \, cm$ and the last term $l = 25 \, cm$.
The total length of the wood required is the sum of the $A.P.$ series,given by the formula $S_n = \frac{n}{2}(a + l)$.
Substituting the values: $S_{11} = \frac{11}{2}(45 + 25) = \frac{11}{2}(70) = 11 \times 35 = 385 \, cm$.
Thus,the total length of wood required for the rungs is $385 \, cm$.

(A) The house numbers are $1, 2, 3, \ldots, 49$. This forms an Arithmetic Progression $(A.P.)$ with first term $a = 1$ and common difference $d = 1$.
Let the house number be $x$. The sum of the house numbers preceding $x$ is the sum of the first $(x-1)$ terms: $S_{x-1} = \frac{(x-1)}{2}[2a + (x-1-1)d] = \frac{x-1}{2}[2(1) + (x-2)(1)] = \frac{x(x-1)}{2}$.
The sum of the house numbers following $x$ is the total sum of $49$ houses minus the sum of the first $x$ houses: $S_{49} - S_x = \frac{49}{2}[2(1) + (49-1)(1)] - \frac{x}{2}[2(1) + (x-1)(1)] = \frac{49 \times 50}{2} - \frac{x(x+1)}{2} = 1225 - \frac{x(x+1)}{2}$.
Equating the two sums: $\frac{x(x-1)}{2} = 1225 - \frac{x(x+1)}{2}$.
Multiplying by $2$: $x^2 - x = 2450 - (x^2 + x)$.
$x^2 - x = 2450 - x^2 - x$.
$2x^2 = 2450$.
$x^2 = 1225$.
$x = \sqrt{1225} = 35$ (since $x$ must be positive).
Thus,the value of $x$ is $35$.

(B) From the figure,it can be observed that:
$1^{\text{st}}$ step is $\frac{1}{2} \, m$ wide,
$2^{\text{nd}}$ step is $1 \, m$ wide,
$3^{\text{rd}}$ step is $\frac{3}{2} \, m$ wide.
Therefore,the width of each step is increasing by $\frac{1}{2} \, m$ each time,whereas their height $\frac{1}{4} \, m$ and length $50 \, m$ remain the same.
Thus,the widths of these steps form an arithmetic progression: $\frac{1}{2}, 1, \frac{3}{2}, 2, \dots$
Volume of concrete in $1^{\text{st}}$ step $= \text{length} \times \text{width} \times \text{height} = 50 \times \frac{1}{2} \times \frac{1}{4} = \frac{25}{4} \, m^3$.
Volume of concrete in $2^{\text{nd}}$ step $= 50 \times 1 \times \frac{1}{4} = \frac{25}{2} \, m^3$.
Volume of concrete in $3^{\text{rd}}$ step $= 50 \times \frac{3}{2} \times \frac{1}{4} = \frac{75}{4} \, m^3$.
The volumes of concrete in these steps form an $A.P.$ with first term $a = \frac{25}{4}$ and common difference $d = \frac{25}{2} - \frac{25}{4} = \frac{25}{4}$.
The total volume for $n = 15$ steps is given by the sum $S_n = \frac{n}{2} [2a + (n - 1)d]$.
$S_{15} = \frac{15}{2} \left[ 2 \left( \frac{25}{4} \right) + (15 - 1) \left( \frac{25}{4} \right) \right]$
$S_{15} = \frac{15}{2} \left[ \frac{25}{2} + 14 \times \frac{25}{4} \right] = \frac{15}{2} \left[ \frac{25}{2} + \frac{175}{2} \right]$
$S_{15} = \frac{15}{2} \left[ \frac{200}{2} \right] = \frac{15}{2} \times 100 = 750$.
Thus,the total volume of concrete required is $750 \, m^3$.