Solution:

Sum of the following APs:

(i) 2, 7, 12, ..., to 10 terms:

First term, $a = 2$, common difference, $d = 7 - 2 = 5$.

Number of terms, $n = 10$.

Sum of n terms, $S_n = \dfrac{n}{2} \left( 2a + (n-1) d \right)$.

Substitute the values: $S_{10} = \dfrac{10}{2} \left( 2(2) + (10-1)(5) \right) = 5 \left( 4 + 45 \right) = 5 \times 49 = 245$.

(ii) –37, –33, –29, ..., to 12 terms:

First term, $a = -37$, common difference, $d = -33 - (-37) = 4$.

Number of terms, $n = 12$.

Sum of n terms, $S_n = \dfrac{n}{2} \left( 2a + (n-1) d \right)$.

Substitute the values: $S_{12} = \dfrac{12}{2} \left( 2(-37) + (12-1)(4) \right) = 6 \left( -74 + 44 \right) = 6 \times (-30) = -180$.

(iii) 0.6, 1.7, 2.8, ..., to 100 terms:

First term, $a = 0.6$, common difference, $d = 1.7 - 0.6 = 1.1$.

Number of terms, $n = 100$.

Sum of n terms, $S_n = \dfrac{n}{2} \left( 2a + (n-1) d \right)$.

Substitute the values: $S_{100} = \dfrac{100}{2} \left( 2(0.6) + (100-1)(1.1) \right) = 50 \left( 1.2 + 99 \times 1.1 \right) = 50 \left( 1.2 + 108.9 \right) = 50 \times 110.1 = 5505$.

(iv) $\dfrac{1}{15}$, $\dfrac{1}{12}$, $\dfrac{1}{10}$,..., to 11 terms:

First term, $a = \dfrac{1}{15}$, common difference, $d = \dfrac{1}{12} - \dfrac{1}{15} = \dfrac{5 - 4}{60} = \dfrac{1}{60}$.

Number of terms, $n = 11$.

Sum of n terms, $S_n = \dfrac{n}{2} \left( 2a + (n-1) d \right)$.

Substitute the values: $S_{11} = \dfrac{11}{2} \left( 2\left( \dfrac{1}{15} \right) + (11-1) \left( \dfrac{1}{60} \right) \right)$.

Calculate the terms: $S_{11} = \dfrac{11}{2} \left( \dfrac{2}{15} + 10 \times \dfrac{1}{60} \right) = \dfrac{11}{2} \left( \dfrac{2}{15} + \dfrac{10}{60} \right) = \dfrac{11}{2} \left( \dfrac{2}{15} + \dfrac{1}{6} \right)$.

Find LCM of 15 and 6: LCM = 30.

$S_{11} = \dfrac{11}{2} \left( \dfrac{4}{30} + \dfrac{5}{30} \right) = \dfrac{11}{2} \times \dfrac{9}{30} = \dfrac{11 \times 9}{2 \times 30} = \dfrac{99}{60} = \dfrac{33}{20}$.

Solution:

(i) 7 + 10$\dfrac{1}{2}$ + 14 + ... + 84

(ii) 34 + 32 + 30 + ... + 10

(iii) –5 + (–8) + (–11) + ... + (–230)

Solution:

(i) Given a=5, d=3, a_n = 50, find n and S$_n$.

Formula for nth term of an AP: $a_n = a + (n-1) \cdot d$

Substitute the known values: $50 = 5 + (n-1) \cdot 3$

Solving for n: $50 - 5 = (n-1) \cdot 3$

$45 = (n-1) \cdot 3$

$n - 1 = 15$

$n = 16$

Now, for sum of n terms: $S_n = \frac{n}{2} \cdot (a + a_n)$

$S_{16} = \frac{16}{2} \cdot (5 + 50)$

$S_{16} = 8 \cdot 55 = 440$

(ii) Given a = 7, a$_{13}$ = 35, find d and S$_{13}$.

Formula for nth term of an AP: $a_n = a + (n-1) \cdot d$

Substitute the known values: $35 = 7 + (13-1) \cdot d$

$35 - 7 = 12 \cdot d$

$28 = 12 \cdot d$

$d = \frac{28}{12} = \frac{7}{3}$

Now, for sum of n terms: $S_n = \frac{n}{2} \cdot (2a + (n-1) \cdot d)$

$S_{13} = \frac{13}{2} \cdot (2 \cdot 7 + (13-1) \cdot \frac{7}{3})$

$S_{13} = \frac{13}{2} \cdot (14 + 12 \cdot \frac{7}{3})$

$S_{13} = \frac{13}{2} \cdot \left(14 + 28\right)$

$S_{13} = \frac{13}{2} \cdot 42 = 13 \cdot 21 = 273$

(iii) Given a$_{12}$ = 37, d = 3, find a and S$_{12}$.

Formula for nth term of an AP: $a_n = a + (n-1) \cdot d$

Substitute the known values: $37 = a + (12-1) \cdot 3$

$37 = a + 33$

$a = 37 - 33 = 4$

Now, for sum of n terms: $S_n = \frac{n}{2} \cdot (2a + (n-1) \cdot d)$

$S_{12} = \frac{12}{2} \cdot (2 \cdot 4 + (12-1) \cdot 3)$

$S_{12} = 6 \cdot (8 + 33)$

$S_{12} = 6 \cdot 41 = 246$

(iv) Given a$_{3}$ = 15, S$_{10}$ = 125, find d and a$_{10}$.

Formula for nth term of an AP: $a_n = a + (n-1) \cdot d$

Substitute the known value for $a_3$: $15 = a + (3-1) \cdot d$

$15 = a + 2 \cdot d$

$a = 15 - 2 \cdot d$

Now, for sum of n terms: $S_n = \frac{n}{2} \cdot (2a + (n-1) \cdot d)$

$S_{10} = \frac{10}{2} \cdot (2a + (10-1) \cdot d)$

$125 = 5 \cdot (2a + 9 \cdot d)$

$25 = 2a + 9d$

Substitute $a = 15 - 2d$: $25 = 2(15 - 2d) + 9d$

$25 = 30 - 4d + 9d$

$25 = 30 + 5d$

$-5 = 5d$

$d = -1$

Now, find $a_{10}$: $a_{10} = a + (10-1) \cdot d = a + 9 \cdot (-1) = a - 9$

$a_{10} = (15 - 2 \cdot (-1)) - 9 = 15 + 2 - 9 = 8$

(v) Given d = 5, S$_9$ = 75, find a and a$_9$.

For sum of n terms: $S_n = \frac{n}{2} \cdot (2a + (n-1) \cdot d)$

$75 = \frac{9}{2} \cdot (2a + 8 \cdot 5)$

$75 = \frac{9}{2} \cdot (2a + 40)$

$75 \cdot 2 = 9 \cdot (2a + 40)$

$150 = 9 \cdot (2a + 40)$

$\frac{150}{9} = 2a + 40$

$\frac{50}{3} = 2a + 40$

$2a = \frac{50}{3} - 40 = \frac{50}{3} - \frac{120}{3} = \frac{-70}{3}$

$a = \frac{-35}{3}$

Now, find $a_9$: $a_9 = a + (9-1) \cdot d = a + 8 \cdot 5 = a + 40$

$a_9 = \frac{-35}{3} + 40 = \frac{-35}{3} + \frac{120}{3} = \frac{85}{3}$

(vi) Given a = 2, d = 8, S$_n$ = 90, find n and a$_n$.

For sum of n terms: $S_n = \frac{n}{2} \cdot (2a + (n-1) \cdot d)$

$90 = \frac{n}{2} \cdot (2 \cdot 2 + (n-1) \cdot 8)$

$90 = \frac{n}{2} \cdot (4 + 8n - 8)$

$90 = \frac{n}{2} \cdot (8n - 4)$

$90 = n \cdot (4n - 2)$

$90 = 4n^2 - 2n$

$4n^2 - 2n - 90 = 0$

Solving the quadratic equation, $n = 5$ (discarding negative root)

Now, for $a_n$: $a_n = a + (n-1) \cdot d = 2 + (5-1) \cdot 8 = 2 + 32 = 34$

(vii) Given a = 8, a$_n$ = 62, S$_n$ = 210, find n and d.

Formula for nth term of an AP: $a_n = a + (n-1) \cdot d$

$62 = 8 + (n-1) \cdot d$

$62 - 8 = (n-1) \cdot d$

$54 = (n-1) \cdot d$

$d = \frac{54}{n-1}$

Now, for sum of n terms: $S_n = \frac{n}{2} \cdot (a + a_n)$

$210 = \frac{n}{2} \cdot (8 + 62)$

$210 = \frac{n}{2} \cdot 70$

$210 = 35n$

$n = 6$

Now, find $d$: $d = \frac{54}{6-1} = \frac{54}{5} = 10.8$

(viii) Given a$_n$ = 4, d = 2, S$_n$ = –14, find n and a.

Formula for nth term of an AP: $a_n = a + (n-1) \cdot d$

$4 = a + (n-1) \cdot 2$

$4 = a + 2n - 2$

$a = 6 - 2n$

Now, for sum of n terms: $S_n = \frac{n}{2} \cdot (2a + (n-1) \cdot d)$

$-14 = \frac{n}{2} \cdot (2(6 - 2n) + (n-1) \cdot 2)$

$-14 = \frac{n}{2} \cdot (12 - 4n + 2n - 2)$

$-14 = \frac{n}{2} \cdot (10 - 2n)$

$-28 = n \cdot (10 - 2n)$

$-28 = 10n - 2n^2$

$2n^2 - 10n - 28 = 0$

Solving this quadratic equation, $n = 7$ (discarding negative root)

Now, find $a$: $a = 6 - 2 \cdot 7 = 6 - 14 = -8$

(ix) Given a = 3, n = 8, S = 192, find d.

For sum of n terms: $S_n = \frac{n}{2} \cdot (2a + (n-1) \cdot d)$

$192 = \frac{8}{2} \cdot (2 \cdot 3 + (8-1) \cdot d)$

$192 = 4 \cdot (6 + 7d)$

$192 = 24 + 28d$

$168 = 28d$

$d = \frac{168}{28} = 6$

Solution:

How many terms of the AP: 9, 17, 25,... must be taken to give a sum of 636?

Given: The first term $a = 9$, the common difference $d = 17 - 9 = 8$, and the sum $S_n = 636$.

We use the formula for the sum of the first $n$ terms of an AP:

$S_n = \dfrac{n}{2} [2a + (n - 1)d]$

Substituting the given values into the formula:

$636 = \dfrac{n}{2} [2(9) + (n - 1)(8)]$

Simplifying the equation:

$636 = \dfrac{n}{2} [18 + 8n - 8]$

$636 = \dfrac{n}{2} [8n + 10]$

Multiply both sides by 2 to eliminate the fraction:

$1272 = n(8n + 10)$

Expanding the equation:

$1272 = 8n^2 + 10n$

Rearranging the equation:

$8n^2 + 10n - 1272 = 0$

This is a quadratic equation in $n$. We solve it using the quadratic formula:

$n = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the equation $8n^2 + 10n - 1272 = 0$, the values of $a$, $b$, and $c$ are $a = 8$, $b = 10$, and $c = -1272$.

Substitute these values into the quadratic formula:

$n = \dfrac{-10 \pm \sqrt{10^2 - 4(8)(-1272)}}{2(8)}$

$n = \dfrac{-10 \pm \sqrt{100 + 40608}}{16}$

$n = \dfrac{-10 \pm \sqrt{40708}}{16}$

$n = \dfrac{-10 \pm 202}{16}$

We now have two possible values for $n$:

$n = \dfrac{-10 + 202}{16} = \dfrac{192}{16} = 12$

or

$n = \dfrac{-10 - 202}{16} = \dfrac{-212}{16} = -13.25$ (which is not a valid solution, as $n$ must be a positive integer).

Thus, the number of terms is $n = 12$.

Solution:

Given:

• First term ($a = 5$)
• Last term ($l = 45$)
• Sum of terms ($S_n = 400$)

Formula for sum of an AP:

The sum of the first $n$ terms of an AP is given by:

$S_n = \frac{n}{2} \times (a + l)$

Substitute the known values into the formula:

$400 = \frac{n}{2} \times (5 + 45)$

$400 = \frac{n}{2} \times 50$

$400 = 25n$

$n = \frac{400}{25}$

$n = 16$

Now, to find the common difference $d$:

The $n$-th term of an AP is given by:

$l = a + (n - 1) \times d$

Substitute the known values:

$45 = 5 + (16 - 1) \times d$

$45 = 5 + 15d$

$40 = 15d$

$d = \frac{40}{15}$

$d = \frac{8}{3}$

Answer:

• Number of terms ($n = 16$)
• Common difference ($d = \frac{8}{3}$)

Solution:

Given:

The first term of the AP is $a = 17$, the last term is $l = 350$, and the common difference is $d = 9$.

To find the number of terms ($n$) in the AP:

We use the formula for the $n$-th term of an AP:

$l = a + (n - 1) \cdot d$

Substituting the given values:

$350 = 17 + (n - 1) \cdot 9$

$350 - 17 = (n - 1) \cdot 9$

$333 = (n - 1) \cdot 9$

$n - 1 = \frac{333}{9}$

$n - 1 = 37$

$n = 38$

So, the number of terms is $n = 38$.

To find the sum of the AP:

The sum of the first $n$ terms of an AP is given by:

$S_n = \frac{n}{2} \cdot (a + l)$

Substituting the values of $n$, $a$, and $l$:

$S_{38} = \frac{38}{2} \cdot (17 + 350)$

$S_{38} = 19 \cdot 367$

$S_{38} = 6983$

So, the sum of the terms is $S = 6983$.

Solution:

Given:

• Common difference, $d = 7$
• 22nd term, $a_{22} = 149$

Formula for nth term of an AP:

The nth term of an AP is given by:

$a_n = a + (n - 1) \cdot d$

Finding the first term:

For the 22nd term, $a_{22} = a + (22 - 1) \cdot 7 = 149$

$a + 21 \cdot 7 = 149$

$a + 147 = 149$

$a = 149 - 147 = 2$

Sum of first n terms of an AP:

The sum of the first n terms of an AP is given by:

$S_n = \frac{n}{2} \cdot [2a + (n - 1) \cdot d]$

Substitute values for the sum of first 22 terms:

$S_{22} = \frac{22}{2} \cdot [2 \cdot 2 + (22 - 1) \cdot 7]$

$S_{22} = 11 \cdot [4 + 21 \cdot 7]$

$S_{22} = 11 \cdot [4 + 147]$

$S_{22} = 11 \cdot 151$

$S_{22} = 1661$

Solution:

Finding the sum of first 51 terms of an AP

Given, the second term of the AP is 14 and the third term is 18.

Let the first term be $a$ and the common difference be $d$.

We know the formula for the $n$-th term of an AP is given by:

$T_n = a + (n-1) \cdot d$

For the second term ($T_2$), we have:

$T_2 = a + (2-1) \cdot d = a + d = 14$

For the third term ($T_3$), we have:

$T_3 = a + (3-1) \cdot d = a + 2d = 18$

We now have the system of equations:

1. $a + d = 14$

2. $a + 2d = 18$

Subtract equation 1 from equation 2:

$(a + 2d) - (a + d) = 18 - 14$

$d = 4$

Now substitute $d = 4$ in equation 1:

$a + 4 = 14$

$a = 10$

Now, we know the first term $a = 10$ and the common difference $d = 4$.

The sum of the first $n$ terms of an AP is given by the formula:

$S_n = \frac{n}{2} \cdot [2a + (n-1) \cdot d]$

For $n = 51$, the sum of the first 51 terms is:

$S_{51} = \frac{51}{2} \cdot [2 \cdot 10 + (51 - 1) \cdot 4]$

$S_{51} = \frac{51}{2} \cdot [20 + 200]$

$S_{51} = \frac{51}{2} \cdot 220$

$S_{51} = 51 \cdot 110$

$S_{51} = 5610$

Solution:

Given:

The sum of first 7 terms of an AP is 49 and the sum of first 17 terms is 289.

Let the first term of the AP be $a$ and the common difference be $d$.

Formula for the sum of the first $n$ terms of an AP:

The sum of the first $n$ terms of an AP is given by the formula:

$S_n = \frac{n}{2} \left( 2a + (n-1) d \right)$

For the first 7 terms, we have:

$S_7 = \frac{7}{2} \left( 2a + 6d \right) = 49$

Multiplying both sides by 2, we get:

$7(2a + 6d) = 98$

Dividing by 7:

$2a + 6d = 14 \ \ \ \ \ \text{(Equation 1)}$

For the first 17 terms, we have:

$S_{17} = \frac{17}{2} \left( 2a + 16d \right) = 289$

Multiplying both sides by 2, we get:

$17(2a + 16d) = 578$

Dividing by 17:

$2a + 16d = 34 \ \ \ \ \ \text{(Equation 2)}$

Solving the system of equations:

From Equation 1: $2a + 6d = 14$

From Equation 2: $2a + 16d = 34$

Subtract Equation 1 from Equation 2:

$(2a + 16d) - (2a + 6d) = 34 - 14$

$10d = 20$

Thus, $d = 2$.

Substitute $d = 2$ into Equation 1:

$2a + 6(2) = 14$

$2a + 12 = 14$

$2a = 2$

$a = 1$

Sum of first $n$ terms:

The sum of the first $n$ terms is given by:

$S_n = \frac{n}{2} \left( 2a + (n-1) d \right)$

Substitute $a = 1$ and $d = 2$:

$S_n = \frac{n}{2} \left( 2(1) + (n-1)(2) \right)$

$S_n = \frac{n}{2} \left( 2 + 2n - 2 \right)$

$S_n = \frac{n}{2} \times 2n$

$S_n = n^2$

Solution:

Show that a1, a2, ..., an, ... form an AP where an is defined as below:

We are given two different expressions for $a_n$:

(i) $a_n = 3 + 4n$

To show that the sequence forms an arithmetic progression (AP), we need to check if the difference between consecutive terms is constant.

Let's compute the first few terms:

For $n = 1$, $a_1 = 3 + 4(1) = 7$

For $n = 2$, $a_2 = 3 + 4(2) = 11$

For $n = 3$, $a_3 = 3 + 4(3) = 15$

The common difference $d = a_2 - a_1 = 11 - 7 = 4$ and $a_3 - a_2 = 15 - 11 = 4$. Since the common difference is constant, the sequence forms an AP with a common difference of 4.

The sum of the first 15 terms of an AP is given by:

$S_n = \frac{n}{2} [2a_1 + (n - 1)d]$

For $n = 15$, $a_1 = 7$ and $d = 4$:

$S_{15} = \frac{15}{2} [2(7) + (15 - 1)(4)] = \frac{15}{2} [14 + 56] = \frac{15}{2} \times 70 = 15 \times 35 = 525$

(ii) $a_n = 9 - 5n$

Similarly, we compute the first few terms:

For $n = 1$, $a_1 = 9 - 5(1) = 4$

For $n = 2$, $a_2 = 9 - 5(2) = -1$

For $n = 3$, $a_3 = 9 - 5(3) = -6$

The common difference $d = a_2 - a_1 = -1 - 4 = -5$ and $a_3 - a_2 = -6 - (-1) = -5$. Since the common difference is constant, the sequence forms an AP with a common difference of -5.

The sum of the first 15 terms of this AP is:

$S_n = \frac{n}{2} [2a_1 + (n - 1)d]$

For $n = 15$, $a_1 = 4$ and $d = -5$:

$S_{15} = \frac{15}{2} [2(4) + (15 - 1)(-5)] = \frac{15}{2} [8 + (-70)] = \frac{15}{2} \times (-62) = 15 \times (-31) = -465$

Solution:

Given: The sum of the first n terms of an AP is S_n = 4n – n$^2$.

1. First term (S₁):

To find the first term, substitute n = 1 in the sum formula:

S₁ = 4(1) – (1)$^2$ = 4 – 1 = 3

The first term is 3.

2. Sum of the first two terms:

To find the sum of the first two terms, substitute n = 2 in the sum formula:

S₂ = 4(2) – (2)$^2$ = 8 – 4 = 4

The sum of the first two terms is 4.

3. Second term:

The second term can be found by subtracting the sum of the first term from the sum of the first two terms:

Second term = S₂ – S₁ = 4 – 3 = 1

The second term is 1.

4. Third term:

To find the sum of the first three terms, substitute n = 3 in the sum formula:

S₃ = 4(3) – (3)$^2$ = 12 – 9 = 3

The third term = S₃ – S₂ = 3 – 4 = -1

The third term is -1.

5. Tenth term:

To find the sum of the first ten terms, substitute n = 10 in the sum formula:

S₁₀ = 4(10) – (10)$^2$ = 40 – 100 = -60

The tenth term = S₁₀ – S₉.

First, calculate S₉: S₉ = 4(9) – (9)$^2$ = 36 – 81 = -45.

The tenth term = -60 – (-45) = -60 + 45 = -15.

The tenth term is -15.

6. nth term:

The nth term can be found using the formula:

a_n = S_n – S_n-1.

Substitute the expressions for S_n and S_n-1: a_n = (4n – n$^2$) – (4(n-1) – (n-1)$^2$).

Simplifying the expression will give the nth term.

Solution:

Sum of the first 40 positive integers divisible by 6

The integers divisible by 6 are in the form of $6n$, where $n$ is a positive integer. The first 40 integers divisible by 6 are:

$6, 12, 18, 24, ..., 6 \times 40$

This is an arithmetic progression (AP) where:

• First term $a = 6$
• Common difference $d = 6$
• Number of terms $n = 40$

The sum of the first $n$ terms of an AP is given by the formula:

$S_n = \dfrac{n}{2} \times (2a + (n-1) \times d)$

Substitute the values:

$S_{40} = \dfrac{40}{2} \times (2 \times 6 + (40-1) \times 6)$

$S_{40} = 20 \times (12 + 39 \times 6)$

$S_{40} = 20 \times (12 + 234)$

$S_{40} = 20 \times 246$

$S_{40} = 4920$

The sum of the first 40 positive integers divisible by 6 is $4920$.

Solution:

Find the sum of the first 15 multiples of 8

We are asked to find the sum of the first 15 multiples of 8. These multiples are given by:

$ 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120 $

This is an arithmetic sequence where:

• The first term $a = 8$
• The common difference $d = 8$
• The number of terms $n = 15$

The sum $S_n$ of the first $n$ terms of an arithmetic sequence is given by the formula:

$ S_n = \frac{n}{2} \times (2a + (n-1) \times d) $

Substitute the values:

$ S_{15} = \frac{15}{2} \times (2 \times 8 + (15-1) \times 8) $

$ S_{15} = \frac{15}{2} \times (16 + 112) $

$ S_{15} = \frac{15}{2} \times 128 $

$ S_{15} = 15 \times 64 $

$ S_{15} = 960 $

Therefore, the sum of the first 15 multiples of 8 is $ 960 $.

Solution:

Sum of Odd Numbers Between 0 and 50

We are asked to find the sum of all odd numbers between 0 and 50. The odd numbers between 0 and 50 are:

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49

We can observe that this is an arithmetic sequence with the first term $a = 1$, the common difference $d = 2$, and the last term $l = 49$.

To find the sum of the series, we use the formula for the sum of an arithmetic series:

$S_n = \frac{n}{2} \times (a + l)$

Where $n$ is the number of terms. To find $n$, we use the formula:

$n = \frac{l - a}{d} + 1$

Substituting the values:

$n = \frac{49 - 1}{2} + 1 = \frac{48}{2} + 1 = 24 + 1 = 25$

Now, substitute the values of $n$, $a$, and $l$ into the sum formula:

$S_{25} = \frac{25}{2} \times (1 + 49) = \frac{25}{2} \times 50 = 25 \times 25 = 625$

Thus, the sum of the odd numbers between 0 and 50 is $625$.

Solution:

Solution:

Solution:

Problem:

A sum of Rs. 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs. 20 less than its preceding prize, find the value of each of the prizes.

Solution:

Let the value of the first prize be $x$. Then, the value of the second prize is $(x - 20)$, the value of the third prize is $(x - 40)$, and so on.

The value of the seventh prize will be $(x - 120)$.

The total sum of all the prizes is given as Rs. 700. Therefore, we can write the equation as:

$x + (x - 20) + (x - 40) + (x - 60) + (x - 80) + (x - 100) + (x - 120) = 700$

Simplifying the left-hand side:

$7x - (20 + 40 + 60 + 80 + 100 + 120) = 700$

$7x - 420 = 700$

Now, solving for $x$:

$7x = 700 + 420$

$7x = 1120$

$x = \frac{1120}{7}$

$x = 160$

So, the value of the first prize is Rs. 160. The other prizes will be as follows:

• First prize = Rs. 160
• Second prize = Rs. 140
• Third prize = Rs. 120
• Fourth prize = Rs. 100
• Fifth prize = Rs. 80
• Sixth prize = Rs. 60
• Seventh prize = Rs. 40

Solution:

Solution:

We are given that each section of each class will plant a number of trees equal to the class number. There are 3 sections in each class, and the classes range from I to XII.

The number of trees planted by each section of Class I will be 1, for Class II it will be 2, and so on up to Class XII, where each section will plant 12 trees.

The total number of trees planted by the students can be calculated as follows:

Total number of trees = 3 × (1 + 2 + 3 + ... + 12)

We know the sum of the first n natural numbers is given by the formula:

$S_n = \frac{n(n + 1)}{2}$

For n = 12, the sum of the numbers from 1 to 12 is:

$S_{12} = \frac{12(12 + 1)}{2} = \frac{12 \times 13}{2} = 78$

Thus, the total number of trees planted by the students is:

Total number of trees = 3 × 78 = 234

Solution:

Question: Total Length of a Spiral Made up of Thirteen Semicircles

A spiral is made up of successive semicircles with centres alternately at A and B, starting with centre at A, and radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ..., as shown in Fig. 5.4. We are asked to find the total length of such a spiral made up of thirteen consecutive semicircles. (Take $\pi$ = 22/7)

Solution:

The length of a semicircle is given by the formula: $L = \pi \times r$, where $r$ is the radius of the semicircle.

The radii of the semicircles are: 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ..., up to 6.5 cm for the 13th semicircle. We can list these radii as:

• Radius 1: 0.5 cm
• Radius 2: 1.0 cm
• Radius 3: 1.5 cm
• Radius 4: 2.0 cm
• Radius 5: 2.5 cm
• Radius 6: 3.0 cm
• Radius 7: 3.5 cm
• Radius 8: 4.0 cm
• Radius 9: 4.5 cm
• Radius 10: 5.0 cm
• Radius 11: 5.5 cm
• Radius 12: 6.0 cm
• Radius 13: 6.5 cm

The total length of the spiral is the sum of the lengths of these semicircles:

Length of each semicircle = $L = \frac{22}{7} \times r$, where $r$ is the radius.

Total length = $L_1 + L_2 + L_3 + \cdots + L_{13}$

Total length = $\frac{22}{7} \times (0.5 + 1.0 + 1.5 + 2.0 + 2.5 + 3.0 + 3.5 + 4.0 + 4.5 + 5.0 + 5.5 + 6.0 + 6.5)$

Sum of radii = $0.5 + 1.0 + 1.5 + 2.0 + 2.5 + 3.0 + 3.5 + 4.0 + 4.5 + 5.0 + 5.5 + 6.0 + 6.5 = 45.5$

Total length = $\dfrac{22}{7} \times 45.5$

Total length = $\dfrac{22 \times 45.5}{7}$

Total length = 143 cm

Solution:

Solution:

Solution:

Potato Race Total Distance Calculation

In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

The competitor runs back and forth, picking up one potato at a time and returning it to the bucket. The total distance run by the competitor can be calculated for each potato pickup.

For the first potato:

The competitor runs 5 m to the first potato and then 5 m back to the bucket. The total distance for the first potato is:

$2 \times 5 = 10$ m

For the second potato:

The competitor runs 5 m to the second potato, which is 3 m further, and then 5 m + 3 m back to the bucket. The total distance for the second potato is:

$2 \times (5 + 3) = 16$ m

For the third potato:

The competitor runs 5 m + 3 m to the third potato and then 5 m + 3 m back to the bucket. The total distance for the third potato is:

$2 \times (5 + 3 \times 2) = 22$ m

Continuing this pattern for each subsequent potato, the total distance for the $n$-th potato can be expressed as:

$2 \times (5 + 3 \times (n - 1))$

The total distance the competitor runs is the sum of the distances for all 10 potatoes:

Total distance = $2 \times (5 + 3 \times 0) + 2 \times (5 + 3 \times 1) + \ldots + 2 \times (5 + 3 \times 9)$