NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.3#

Here $a = 2$, $n = 10$ and:

$$d = 7 - 2 = 5$$

Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$, we find $S_{10}$:

$$= \dfrac{10}{2}[2(2) + (10-1)(5)]$$

$$= 5[4 + 45]$$

$$= 5 \times 49$$

$$= 245$$

∴ $S_{10} = \mathbf{245}$

Here $a = -37$, $n = 12$ and:

$$d = -33 - (-37) = 4$$

Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$, we find $S_{12}$:

$$= \dfrac{12}{2}[2(-37) + (12-1)(4)]$$

$$= 6[-74 + 44]$$

$$= 6 \times (-30)$$

$$= -180$$

∴ $S_{12} = \mathbf{-180}$

Here $a = 0.6$, $n = 100$ and:

$$d = 1.7 - 0.6 = 1.1$$

Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$, we find $S_{100}$:

$$= \dfrac{100}{2}[2(0.6) + (100-1)(1.1)]$$

$$= 50[1.2 + 108.9]$$

$$= 50 \times 110.1$$

$$= 5505$$

∴ $S_{100} = \mathbf{5505}$

Here $a = \dfrac{1}{15}$, $n = 11$ and:

$$d = \dfrac{1}{12} - \dfrac{1}{15} = \dfrac{5 - 4}{60} = \dfrac{1}{60}$$

Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$, we find $S_{11}$:

$$= \dfrac{11}{2}\left[2 \times \dfrac{1}{15} + (11-1) \times \dfrac{1}{60}\right]$$

$$= \dfrac{11}{2}\left[\dfrac{2}{15} + \dfrac{1}{6}\right]$$

$$= \dfrac{11}{2}\left[\dfrac{4 + 5}{30}\right]$$

$$= \dfrac{11}{2} \times \dfrac{3}{10}$$

$$= \dfrac{33}{20}$$

∴ $S_{11} = \mathbf{\dfrac{33}{20}}$

Here $a = 7$, $l = 84$ and:

$$d = 10\dfrac{1}{2} - 7 = \dfrac{7}{2}$$

Finding $n$ using $a_n = a + (n-1)d$:

$$84 = 7 + (n-1) \times \dfrac{7}{2}$$

$$\Rightarrow \dfrac{7}{2}(n-1) = 77$$

$$\Rightarrow n - 1 = 22$$

$$\Rightarrow n = 23$$

Using $S_n = \dfrac{n}{2}(a + l)$, we find $S_{23}$:

$$= \dfrac{23}{2}(7 + 84)$$

$$= \dfrac{23 \times 91}{2}$$

$$= \dfrac{2093}{2}$$

$$= 1046\dfrac{1}{2}$$

∴ The sum is $\mathbf{1046\dfrac{1}{2}}$.

Here $a = 34$, $l = 10$ and:

$$d = 32 - 34 = -2$$

Finding $n$ using $a_n = a + (n-1)d$:

$$10 = 34 + (n-1)(-2)$$

$$\Rightarrow -2(n-1) = -24$$

$$\Rightarrow n - 1 = 12$$

$$\Rightarrow n = 13$$

Using $S_n = \dfrac{n}{2}(a + l)$, we find $S_{13}$:

$$= \dfrac{13}{2}(34 + 10)$$

$$= \dfrac{13 \times 44}{2}$$

$$= 286$$

∴ The sum is 286.

Here $a = -5$, $l = -230$ and:

$$d = -8 - (-5) = -3$$

Finding $n$ using $a_n = a + (n-1)d$:

$$-230 = -5 + (n-1)(-3)$$

$$\Rightarrow -3(n-1) = -225$$

$$\Rightarrow n - 1 = 75$$

$$\Rightarrow n = 76$$

Using $S_n = \dfrac{n}{2}(a + l)$, we find $S_{76}$:

$$= \dfrac{76}{2}[-5 + (-230)]$$

$$= 38 \times (-235)$$

$$= -8930$$

∴ The sum is −8930.

Using $a_n = a + (n-1)d$:

$$50 = 5 + (n-1)(3)$$

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

$$\Rightarrow n - 1 = 15$$

$$\Rightarrow n = 16$$

Using $S_n = \dfrac{n}{2}(a + a_n)$, we find $S_{16}$:

$$= \dfrac{16}{2}(5 + 50)$$

$$= 8 \times 55$$

$$= 440$$

∴ $n = \mathbf{16}$ and $S_n = \mathbf{440}$

Using $a_n = a + (n-1)d$:

$$35 = 7 + 12d$$

$$\Rightarrow 12d = 28$$

$$\Rightarrow d = \dfrac{7}{3}$$

Using $S_n = \dfrac{n}{2}(a + a_n)$, we find $S_{13}$:

$$= \dfrac{13}{2}(7 + 35)$$

$$= \dfrac{13 \times 42}{2}$$

$$= 273$$

∴ $d = \mathbf{\dfrac{7}{3}}$ and $S_{13} = \mathbf{273}$

Using $a_n = a + (n-1)d$:

$$37 = a + 11(3)$$

$$\Rightarrow 37 = a + 33$$

$$\Rightarrow a = 4$$

Using $S_n = \dfrac{n}{2}(a + a_n)$, we find $S_{12}$:

$$= \dfrac{12}{2}(4 + 37)$$

$$= 6 \times 41$$

$$= 246$$

∴ $a = \mathbf{4}$ and $S_{12} = \mathbf{246}$

The 3rd term is 15, so:

$$a + 2d = 15 \quad \cdots (1)$$

The sum of the first 10 terms is 125, so:

$$\dfrac{10}{2}[2a + 9d] = 125$$

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

$$\Rightarrow 2a + 9d = 25 \quad \cdots (2)$$

Multiplying equation (1) by 2:

$$2a + 4d = 30 \quad \cdots (3)$$

Subtracting equation (3) from equation (2):

$$\begin{array}{rcl} 2a + 9d &=& 25 \\ 2a + 4d &=& 30 \\ \hline 5d &=& -5 \end{array}$$

$$\Rightarrow d = -1$$

Substituting $d = -1$ in equation (1):

$$a + 2d = 15$$

$$a + 2(-1) = 15$$

$$\Rightarrow a = 17$$

Using $a_n = a + (n-1)d$:

$$a_{10} = 17 + 9(-1) = 17 - 9 = 8$$

∴ $d = \mathbf{-1}$ and $a_{10} = \mathbf{8}$

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

$$75 = \dfrac{9}{2}[2a + 8(5)]$$

$$\Rightarrow 75 = \dfrac{9}{2}(2a + 40)$$

$$\Rightarrow 75 = 9(a + 20)$$

$$\Rightarrow a + 20 = \dfrac{25}{3}$$

$$\Rightarrow a = -\dfrac{35}{3}$$

Using $a_n = a + (n-1)d$, we find $a_9$:

$$= -\dfrac{35}{3} + 8(5)$$

$$= \dfrac{-35 + 120}{3}$$

$$= \dfrac{85}{3}$$

∴ $a = \mathbf{-\dfrac{35}{3}}$ and $a_9 = \mathbf{\dfrac{85}{3}}$

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

$$90 = \dfrac{n}{2}[2(2) + (n-1)(8)]$$

$$\Rightarrow 90 = \dfrac{n}{2}(8n - 4)$$

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

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

$$\Rightarrow 2n^2 - n - 45 = 0$$

We factorise $2n^2 - n - 45$:

$$= 2n^2 - 10n + 9n - 45$$

$$= 2n(n - 5) + 9(n - 5)$$

$$= (n - 5)(2n + 9)$$

So:

$$(n - 5)(2n + 9) = 0$$

$$\Rightarrow n = 5 \text{ or } n = -\dfrac{9}{2}$$

$n$ must be a positive integer, so $n = 5$.

Using $a_n = a + (n-1)d$:

$$a_5 = 2 + 4(8) = 2 + 32 = 34$$

∴ $n = \mathbf{5}$ and $a_n = \mathbf{34}$

Using $S_n = \dfrac{n}{2}(a + a_n)$:

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

$$\Rightarrow 210 = 35n$$

$$\Rightarrow n = 6$$

Using $a_n = a + (n-1)d$:

$$62 = 8 + (6-1)d$$

$$\Rightarrow 5d = 54$$

$$\Rightarrow d = \dfrac{54}{5}$$

∴ $n = \mathbf{6}$ and $d = \mathbf{\dfrac{54}{5}}$

The $n$th term is 4, so:

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

$$\Rightarrow a = 6 - 2n \quad \cdots (1)$$

The sum of $n$ terms is −14, so:

$$\dfrac{n}{2}(a + 4) = -14$$

$$\Rightarrow n(a + 4) = -28 \quad \cdots (2)$$

Substituting $a = 6 - 2n$ in equation (2):

$$n(a + 4) = -28$$

$$n(6 - 2n + 4) = -28$$

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

$$\Rightarrow n^2 - 5n - 14 = 0$$

We factorise $n^2 - 5n - 14$:

$$= n^2 - 7n + 2n - 14$$

$$= n(n - 7) + 2(n - 7)$$

$$= (n - 7)(n + 2)$$

So:

$$(n - 7)(n + 2) = 0$$

$$\Rightarrow n = 7 \text{ or } n = -2$$

$n$ must be a positive integer, so $n = 7$.

Substituting $n = 7$ in equation (1):

$$a = 6 - 2n$$

$$a = 6 - 2(7)$$

$$\Rightarrow a = -8$$

∴ $n = \mathbf{7}$ and $a = \mathbf{-8}$

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

$$192 = \dfrac{8}{2}[2(3) + 7d]$$

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

$$\Rightarrow 6 + 7d = 48$$

$$\Rightarrow 7d = 42$$

$$\Rightarrow d = 6$$

∴ $d = \mathbf{6}$

Using $S_n = \dfrac{n}{2}(a + l)$:

$$144 = \dfrac{9}{2}(a + 28)$$

$$\Rightarrow a + 28 = 32$$

$$\Rightarrow a = 4$$

∴ $a = \mathbf{4}$

Here $a = 9$, $S_n = 636$ and:

$$d = 17 - 9 = 8$$

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

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

$$\Rightarrow 636 = \dfrac{n}{2}(8n + 10)$$

$$\Rightarrow 636 = n(4n + 5)$$

$$\Rightarrow 4n^2 + 5n - 636 = 0$$

We factorise $4n^2 + 5n - 636$:

$$= 4n^2 + 53n - 48n - 636$$

$$= n(4n + 53) - 12(4n + 53)$$

$$= (4n + 53)(n - 12)$$

So:

$$(4n + 53)(n - 12) = 0$$

$$\Rightarrow n = 12 \text{ or } n = -\dfrac{53}{4}$$

$n$ must be a positive integer, so $n = 12$.

∴ 12 terms must be taken.

Here $a = 5$, $l = 45$ and $S_n = 400$.

Using $S_n = \dfrac{n}{2}(a + l)$:

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

$$\Rightarrow 400 = 25n$$

$$\Rightarrow n = 16$$

Using $l = a + (n-1)d$:

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

$$\Rightarrow 15d = 40$$

$$\Rightarrow d = \dfrac{8}{3}$$

∴ Number of terms $= \mathbf{16}$ and common difference $= \mathbf{\dfrac{8}{3}}$

Here $a = 17$, $l = 350$ and $d = 9$.

Using $l = a + (n-1)d$:

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

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

$$\Rightarrow n - 1 = 37$$

$$\Rightarrow n = 38$$

Using $S_n = \dfrac{n}{2}(a + l)$, we find $S_{38}$:

$$= \dfrac{38}{2}(17 + 350)$$

$$= 19 \times 367$$

$$= 6973$$

∴ There are 38 terms and their sum is 6973.

Using $a_n = a + (n-1)d$:

$$149 = a + 21(7)$$

$$\Rightarrow 149 = a + 147$$

$$\Rightarrow a = 2$$

Using $S_n = \dfrac{n}{2}(a + l)$, we find $S_{22}$:

$$= \dfrac{22}{2}(2 + 149)$$

$$= 11 \times 151$$

$$= 1661$$

∴ The sum of the first 22 terms is 1661.

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

The second term is 14, so:

$$a + d = 14 \quad \cdots (1)$$

The third term is 18, so:

$$a + 2d = 18 \quad \cdots (2)$$

Subtracting equation (1) from equation (2):

$$\begin{array}{rcl} a + 2d &=& 18 \\ a + d &=& 14 \\ \hline d &=& 4 \end{array}$$

Substituting $d = 4$ in equation (1):

$$a + d = 14$$

$$a + 4 = 14$$

$$\Rightarrow a = 10$$

Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$, we find $S_{51}$:

$$= \dfrac{51}{2}[2(10) + 50(4)]$$

$$= \dfrac{51}{2} \times 220$$

$$= 51 \times 110$$

$$= 5610$$

∴ The sum of the first 51 terms is 5610.

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

The sum of the first 7 terms is 49, so:

$$\dfrac{7}{2}[2a + 6d] = 49$$

$$\Rightarrow 7(a + 3d) = 49$$

$$\Rightarrow a + 3d = 7 \quad \cdots (1)$$

The sum of the first 17 terms is 289, so:

$$\dfrac{17}{2}[2a + 16d] = 289$$

$$\Rightarrow 17(a + 8d) = 289$$

$$\Rightarrow a + 8d = 17 \quad \cdots (2)$$

Subtracting equation (1) from equation (2):

$$\begin{array}{rcl} a + 8d &=& 17 \\ a + 3d &=& 7 \\ \hline 5d &=& 10 \end{array}$$

$$\Rightarrow d = 2$$

Substituting $d = 2$ in equation (1):

$$a + 3d = 7$$

$$a + 3(2) = 7$$

$$\Rightarrow a = 1$$

Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$, we find $S_n$:

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

$$= \dfrac{n}{2} \times 2n$$

$$= n^2$$

∴ The sum of the first $n$ terms is $\mathbf{n^2}$.

First few terms:
$a_1 = 3 + 4(1) = 7$
$a_2 = 3 + 4(2) = 11$
$a_3 = 3 + 4(3) = 15$

We find $a_{n+1} - a_n$:

$$= [3 + 4(n+1)] - (3 + 4n)$$

$$= 3 + 4n + 4 - 3 - 4n$$

$$= 4$$$$\Rightarrow a_{n+1} - a_n = 4 \text{ for every } n$$

∴ The difference is constant, so the terms form an AP with $a = 7$ and $d = 4$.

Last term:

$$a_{15} = 3 + 4(15) = 63$$

Using $S_n = \dfrac{n}{2}(a + l)$, we find $S_{15}$:

$$= \dfrac{15}{2}(7 + 63)$$

$$= \dfrac{15}{2} \times 70$$

$$= 525$$

∴ $S_{15} = \mathbf{525}$

First few terms:
$a_1 = 9 - 5(1) = 4$
$a_2 = 9 - 5(2) = -1$
$a_3 = 9 - 5(3) = -6$

We find $a_{n+1} - a_n$:

$$= [9 - 5(n+1)] - (9 - 5n)$$

$$= 9 - 5n - 5 - 9 + 5n$$

$$= -5$$$$\Rightarrow a_{n+1} - a_n = -5 \text{ for every } n$$

∴ The difference is constant, so the terms form an AP with $a = 4$ and $d = -5$.

Last term:

$$a_{15} = 9 - 5(15) = -66$$

Using $S_n = \dfrac{n}{2}(a + l)$, we find $S_{15}$:

$$= \dfrac{15}{2}[4 + (-66)]$$

$$= \dfrac{15}{2} \times (-62)$$

$$= -465$$

∴ $S_{15} = \mathbf{-465}$

Given $S_n = 4n - n^2$.

First term:

$$S_1 = 4(1) - 1^2 = 3$$

Sum of the first two terms:

$$S_2 = 4(2) - 2^2 = 4$$

Second term:

$$a_2 = S_2 - S_1 = 4 - 3 = 1$$

Third term:

$$S_3 = 4(3) - 3^2 = 3$$

$$a_3 = S_3 - S_2 = 3 - 4 = -1$$

10th term:

$$S_{10} = 4(10) - 10^2 = -60$$

$$S_9 = 4(9) - 9^2 = -45$$

$$a_{10} = S_{10} - S_9 = -60 - (-45) = -15$$

$n$th term: we find $a_n = S_n - S_{n-1}$:

$$= (4n - n^2) - [4(n-1) - (n-1)^2]$$

$$= 4n - n^2 - 4n + 4 + n^2 - 2n + 1$$

$$= 5 - 2n$$

∴ $S_1 = \mathbf{3}$, $S_2 = \mathbf{4}$, $a_2 = \mathbf{1}$, $a_3 = \mathbf{-1}$, $a_{10} = \mathbf{-15}$ and $a_n = \mathbf{5 - 2n}$

The positive integers divisible by 6 are: $6, 12, 18, \ldots$

Here $a = 6$, $d = 6$ and $n = 40$.

Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$, we find $S_{40}$:

$$= \dfrac{40}{2}[2(6) + 39(6)]$$

$$= 20[12 + 234]$$

$$= 20 \times 246$$

$$= 4920$$

∴ The sum is 4920.

The multiples of 8 are: $8, 16, 24, \ldots$

Here $a = 8$, $d = 8$ and $n = 15$.

Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$, we find $S_{15}$:

$$= \dfrac{15}{2}[2(8) + 14(8)]$$

$$= \dfrac{15}{2} \times 128$$

$$= 15 \times 64$$

$$= 960$$

∴ The sum is 960.

The odd numbers between 0 and 50 are: $1, 3, 5, \ldots, 49$

Here $a = 1$, $d = 2$ and $l = 49$.

Finding $n$ using $a_n = a + (n-1)d$:

$$49 = 1 + (n-1)(2)$$

$$\Rightarrow 2(n-1) = 48$$

$$\Rightarrow n - 1 = 24$$

$$\Rightarrow n = 25$$

Using $S_n = \dfrac{n}{2}(a + l)$, we find $S_{25}$:

$$= \dfrac{25}{2}(1 + 49)$$

$$= \dfrac{25 \times 50}{2}$$

$$= 625$$

∴ The sum is 625.

The penalties form an AP: $200, 250, 300, \ldots$

Here $a = 200$, $d = 50$ and $n = 30$.

Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$, we find $S_{30}$:

$$= \dfrac{30}{2}[2(200) + 29(50)]$$

$$= 15[400 + 1450]$$

$$= 15 \times 1850$$

$$= 27750$$

∴ The contractor has to pay ₹27750 as penalty.

Let the first (highest) prize be ₹$a$. The prizes form an AP with $d = -20$ and $n = 7$.

The seven prizes add up to ₹700, so:

$$\dfrac{7}{2}[2a + 6(-20)] = 700$$

$$\Rightarrow \dfrac{7}{2}(2a - 120) = 700$$

$$\Rightarrow 7(a - 60) = 700$$

$$\Rightarrow a - 60 = 100$$

$$\Rightarrow a = 160$$

∴ The prizes are ₹160, ₹140, ₹120, ₹100, ₹80, ₹60 and ₹40.

Each class has 3 sections, so the trees planted by each class are:
Class I = $3 \times 1 = 3$
Class II = $3 \times 2 = 6$
Class III = $3 \times 3 = 9$

These form an AP: $3, 6, 9, \ldots, 36$, with $a = 3$, $l = 36$ and $n = 12$.

Using $S_n = \dfrac{n}{2}(a + l)$, we find $S_{12}$:

$$= \dfrac{12}{2}(3 + 36)$$

$$= 6 \times 39$$

$$= 234$$

∴ The students will plant 234 trees.

Length of a semicircle of radius $r$ = $\pi r$

So the lengths of the semicircles are $\pi(0.5), \pi(1.0), \pi(1.5), \ldots, \pi(6.5)$.

Total length = $\pi(0.5 + 1.0 + 1.5 + \ldots + 6.5)$

The radii form an AP with $a = 0.5$, $l = 6.5$ and $n = 13$.

Using $S_n = \dfrac{n}{2}(a + l)$, we find $S_{13}$:

$$= \dfrac{13}{2}(0.5 + 6.5)$$

$$= \dfrac{13 \times 7}{2}$$

$$= \dfrac{91}{2}$$

Total length = $\pi \times \dfrac{91}{2}$
Total length = $\dfrac{22}{7} \times \dfrac{91}{2}$
Total length = $143$ cm

∴ The total length of the spiral is 143 cm.

The numbers of logs in the rows form an AP: $20, 19, 18, \ldots$

Here $a = 20$ and $d = -1$. Let there be $n$ rows.

The total number of logs is 200, so:

$$\dfrac{n}{2}[2(20) + (n-1)(-1)] = 200$$

$$\Rightarrow \dfrac{n}{2}(41 - n) = 200$$

$$\Rightarrow 41n - n^2 = 400$$

$$\Rightarrow n^2 - 41n + 400 = 0$$

We factorise $n^2 - 41n + 400$:

$$= n^2 - 16n - 25n + 400$$

$$= n(n - 16) - 25(n - 16)$$

$$= (n - 16)(n - 25)$$

So:

$$(n - 16)(n - 25) = 0$$

$$\Rightarrow n = 16 \text{ or } n = 25$$

For $n = 25$:

$$a_{25} = 20 + 24(-1) = -4$$

The number of logs cannot be negative, so $n = 25$ is not possible.

For $n = 16$:

$$a_{16} = 20 + 15(-1) = 5$$

∴ The logs are placed in 16 rows, with 5 logs in the top row.

The competitor runs to each potato and back, so:
Distance for 1st potato = $2 \times 5 = 10$ m
Distance for 2nd potato = $2 \times (5 + 3) = 16$ m
Distance for 3rd potato = $2 \times (5 + 6) = 22$ m

These distances form an AP: $10, 16, 22, \ldots$, with $a = 10$, $d = 6$ and $n = 10$.

Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$, we find $S_{10}$:

$$= \dfrac{10}{2}[2(10) + 9(6)]$$

$$= 5[20 + 54]$$

$$= 5 \times 74$$

$$= 370$$

∴ The competitor has to run a total distance of 370 m.