NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.2#

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

$$a_8 = 7 + (8-1) \times 3 = 7 + 21 = 28$$

∴ $a_n = \mathbf{28}$

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

$$0 = -18 + (10-1)d$$

$$\Rightarrow 0 = -18 + 9d$$

$$\Rightarrow 9d = 18$$

$$\Rightarrow d = 2$$

∴ $d = \mathbf{2}$

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

$$-5 = a + (18-1)(-3)$$

$$\Rightarrow -5 = a - 51$$

$$\Rightarrow a = -5 + 51$$

$$\Rightarrow a = 46$$

∴ $a = \mathbf{46}$

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

$$3.6 = -18.9 + (n-1)(2.5)$$

$$\Rightarrow 2.5(n-1) = 3.6 + 18.9$$

$$\Rightarrow 2.5(n-1) = 22.5$$

$$\Rightarrow n - 1 = 9$$

$$\Rightarrow n = 10$$

∴ $n = \mathbf{10}$

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

$$a_{105} = 3.5 + (105-1)(0) = 3.5$$

∴ $a_n = \mathbf{3.5}$

Here $a = 10$, $n = 30$ and:

$$d = 7 - 10 = -3$$

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

$$a_{30} = 10 + (30-1)(-3) = 10 - 87 = -77$$

∴ Answer: (C) −77

Here $a = -3$, $n = 11$ and:

$$d = -\dfrac{1}{2} - (-3) = \dfrac{5}{2}$$

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

$$a_{11} = -3 + (11-1)\left(\dfrac{5}{2}\right) = -3 + 25 = 22$$

∴ Answer: (B) 22

Here $a = 2$ and $a_3 = 26$.

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

$$a + 2d = 26$$

$$\Rightarrow 2 + 2d = 26$$

$$\Rightarrow 2d = 24$$

$$\Rightarrow d = 12$$

Missing term $a_2 = 2 + 12 = 14$

∴ The AP is: $2, \mathbf{14}, 26$

Here $a_2 = 13$ and $a_4 = 3$. Using $a_n = a + (n-1)d$:

$$a + d = 13 \quad \cdots (1)$$

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

Subtracting equation (1) from equation (2):

$$\begin{array}{rcl} a + 3d &=& 3 \\ a + d &=& 13 \\ \hline 2d &=& -10 \end{array}$$

$$\Rightarrow d = -5$$

Substituting $d = -5$ in equation (1):

$$a + d = 13$$

$$a + (-5) = 13$$

$$\Rightarrow a = 18$$

Missing term $a_3 = 13 + (-5) = 8$

∴ The AP is: $\mathbf{18}, 13, \mathbf{8}, 3$

Here $a = 5$ and $a_4 = 9\dfrac{1}{2} = \dfrac{19}{2}$.

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

$$a + 3d = \dfrac{19}{2}$$

$$\Rightarrow 5 + 3d = \dfrac{19}{2}$$

$$\Rightarrow 3d = \dfrac{19}{2} - 5$$

$$\Rightarrow 3d = \dfrac{9}{2}$$

$$\Rightarrow d = \dfrac{3}{2}$$

Missing terms:
$a_2 = 5 + \dfrac{3}{2} = \dfrac{13}{2}$
$a_3 = \dfrac{13}{2} + \dfrac{3}{2} = 8$

∴ The AP is: $5, \mathbf{6\dfrac{1}{2}}, \mathbf{8}, 9\dfrac{1}{2}$

Here $a = -4$ and $a_6 = 6$.

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

$$a + 5d = 6$$

$$\Rightarrow -4 + 5d = 6$$

$$\Rightarrow 5d = 10$$

$$\Rightarrow d = 2$$

Missing terms:
$a_2 = -4 + 2 = -2$
$a_3 = -2 + 2 = 0$
$a_4 = 0 + 2 = 2$
$a_5 = 2 + 2 = 4$

∴ The AP is: $-4, \mathbf{-2}, \mathbf{0}, \mathbf{2}, \mathbf{4}, 6$

Here $a_2 = 38$ and $a_6 = -22$. Using $a_n = a + (n-1)d$:

$$a + d = 38 \quad \cdots (1)$$

$$a + 5d = -22 \quad \cdots (2)$$

Subtracting equation (1) from equation (2):

$$\begin{array}{rcl} a + 5d &=& -22 \\ a + d &=& 38 \\ \hline 4d &=& -60 \end{array}$$

$$\Rightarrow d = -15$$

Substituting $d = -15$ in equation (1):

$$a + d = 38$$

$$a + (-15) = 38$$

$$\Rightarrow a = 53$$

Missing terms:
$a_1 = 53$
$a_3 = 38 + (-15) = 23$
$a_4 = 23 + (-15) = 8$
$a_5 = 8 + (-15) = -7$

∴ The AP is: $\mathbf{53}, 38, \mathbf{23}, \mathbf{8}, \mathbf{-7}, -22$

Here $a = 3$, $a_n = 78$ and:

$$d = 8 - 3 = 5$$

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

$$78 = 3 + (n-1) \times 5$$

$$\Rightarrow 5(n-1) = 75$$

$$\Rightarrow n - 1 = 15$$

$$\Rightarrow n = 16$$

∴ 78 is the 16th term of the AP.

Here $a = 7$, $a_n = 205$ and:

$$d = 13 - 7 = 6$$

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

$$205 = 7 + (n-1) \times 6$$

$$\Rightarrow 6(n-1) = 198$$

$$\Rightarrow n - 1 = 33$$

$$\Rightarrow n = 34$$

∴ There are 34 terms.

Here $a = 18$, $a_n = -47$ and:

$$d = 15\dfrac{1}{2} - 18 = -\dfrac{5}{2}$$

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

$$-47 = 18 + (n-1)\left(-\dfrac{5}{2}\right)$$

$$\Rightarrow -\dfrac{5}{2}(n-1) = -65$$

$$\Rightarrow n - 1 = 26$$

$$\Rightarrow n = 27$$

∴ There are 27 terms.

Here $a = 11$ and:

$$d = 8 - 11 = -3$$

Let $-150$ be the $n$th term. Using $a_n = a + (n-1)d$:

$$-150 = 11 + (n-1)(-3)$$

$$\Rightarrow -3(n-1) = -161$$

$$\Rightarrow n - 1 = \dfrac{161}{3}$$

$$\Rightarrow n = \dfrac{164}{3}$$

∴ $n$ is not a positive integer, so $-150$ is not a term of the given AP.

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

The 11th term is 38, so:

$$a + 10d = 38 \quad \cdots (1)$$

The 16th term is 73, so:

$$a + 15d = 73 \quad \cdots (2)$$

Subtracting equation (1) from equation (2):

$$\begin{array}{rcl} a + 15d &=& 73 \\ a + 10d &=& 38 \\ \hline 5d &=& 35 \end{array}$$

$$\Rightarrow d = 7$$

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

$$a + 10d = 38$$

$$a + 10(7) = 38$$

$$\Rightarrow a = -32$$

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

$$a_{31} = -32 + 30(7) = -32 + 210 = 178$$

∴ The 31st term is 178.

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

The 3rd term is 12, so:

$$a + 2d = 12 \quad \cdots (1)$$

The last (50th) term is 106, so:

$$a + 49d = 106 \quad \cdots (2)$$

Subtracting equation (1) from equation (2):

$$\begin{array}{rcl} a + 49d &=& 106 \\ a + 2d &=& 12 \\ \hline 47d &=& 94 \end{array}$$

$$\Rightarrow d = 2$$

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

$$a + 2d = 12$$

$$a + 2(2) = 12$$

$$\Rightarrow a = 8$$

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

$$a_{29} = 8 + 28(2) = 8 + 56 = 64$$

∴ The 29th term is 64.

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

The 3rd term is 4, so:

$$a + 2d = 4 \quad \cdots (1)$$

The 9th term is −8, so:

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

Subtracting equation (1) from equation (2):

$$\begin{array}{rcl} a + 8d &=& -8 \\ a + 2d &=& 4 \\ \hline 6d &=& -12 \end{array}$$

$$\Rightarrow d = -2$$

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

$$a + 2d = 4$$

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

$$\Rightarrow a = 8$$

Let the $n$th term be zero. Using $a_n = a + (n-1)d$:

$$0 = 8 + (n-1)(-2)$$

$$\Rightarrow 0 = 10 - 2n$$

$$\Rightarrow 2n = 10$$

$$\Rightarrow n = 5$$

∴ The 5th term is zero.

The 17th term exceeds the 10th term by 7, so:

$$a_{17} - a_{10} = 7$$

$$\Rightarrow (a + 16d) - (a + 9d) = 7$$

$$\Rightarrow 7d = 7$$

$$\Rightarrow d = 1$$

∴ The common difference is 1.

Here $a = 3$ and:

$$d = 15 - 3 = 12$$

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

$$a_{54} = 3 + 53(12) = 3 + 636 = 639$$

The required term is 132 more than the 54th term, so:

$$a_n = 639 + 132$$

$$\Rightarrow a_n = 771$$

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

$$771 = 3 + (n-1) \times 12$$

$$\Rightarrow 12(n-1) = 768$$

$$\Rightarrow n - 1 = 64$$

$$\Rightarrow n = 65$$

∴ The 65th term is 132 more than the 54th term.

Let the first terms of the two APs be $a$ and $b$, and their common difference be $d$.

The difference between their 100th terms is 100, so:

$$(a + 99d) - (b + 99d) = 100$$

$$\Rightarrow a - b = 100 \quad \cdots (1)$$

Difference between their 1000th terms $= (a + 999d) - (b + 999d)$
$= a - b$
$= 100$ (from equation (1))

∴ The difference between their 1000th terms is also 100.

The three-digit numbers divisible by 7 are: $105, 112, 119, \ldots, 994$

Here $a = 105$, $d = 7$ and $a_n = 994$.

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

$$994 = 105 + (n-1) \times 7$$

$$\Rightarrow 7(n-1) = 889$$

$$\Rightarrow n - 1 = 127$$

$$\Rightarrow n = 128$$

∴ There are 128 three-digit numbers divisible by 7.

The multiples of 4 between 10 and 250 are: $12, 16, 20, \ldots, 248$

Here $a = 12$, $d = 4$ and $a_n = 248$.

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

$$248 = 12 + (n-1) \times 4$$

$$\Rightarrow 4(n-1) = 236$$

$$\Rightarrow n - 1 = 59$$

$$\Rightarrow n = 60$$

∴ There are 60 multiples of 4 between 10 and 250.

For the first AP, $a = 63$ and $d = 2$:

$$a_n = 63 + (n-1)(2) = 2n + 61$$

For the second AP, $a = 3$ and $d = 7$:

$$a_n = 3 + (n-1)(7) = 7n - 4$$

The $n$th terms are equal, so:

$$2n + 61 = 7n - 4$$

$$\Rightarrow 5n = 65$$

$$\Rightarrow n = 13$$

∴ The 13th terms of the two APs are equal.

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

The 7th term exceeds the 5th term by 12, so:

$$a_7 - a_5 = 12$$

$$\Rightarrow (a + 6d) - (a + 4d) = 12$$

$$\Rightarrow 2d = 12$$

$$\Rightarrow d = 6$$

The third term is 16, so:

$$a + 2d = 16$$

Substituting $d = 6$ in this equation:

$$a + 2d = 16$$

$$a + 2(6) = 16$$

$$\Rightarrow a = 4$$

∴ The AP is: $\mathbf{4, 10, 16, 22, \ldots}$

Writing the AP in reverse order: $253, 248, 243, \ldots, 3$

Here $a = 253$ and:

$$d = 248 - 253 = -5$$

The 20th term from the last is the 20th term of this reversed AP. Using $a_n = a + (n-1)d$:

$$a_{20} = 253 + 19(-5) = 253 - 95 = 158$$

∴ The 20th term from the last is 158.

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

The sum of the 4th and 8th terms is 24, so:

$$(a + 3d) + (a + 7d) = 24$$

$$\Rightarrow 2a + 10d = 24$$

$$\Rightarrow a + 5d = 12 \quad \cdots (1)$$

The sum of the 6th and 10th terms is 44, so:

$$(a + 5d) + (a + 9d) = 44$$

$$\Rightarrow 2a + 14d = 44$$

$$\Rightarrow a + 7d = 22 \quad \cdots (2)$$

Subtracting equation (1) from equation (2):

$$\begin{array}{rcl} a + 7d &=& 22 \\ a + 5d &=& 12 \\ \hline 2d &=& 10 \end{array}$$

$$\Rightarrow d = 5$$

Substituting $d = 5$ in equation (1):

$$a + 5d = 12$$

$$a + 5(5) = 12$$

$$\Rightarrow a = -13$$

First three terms:
$a_1 = -13$
$a_2 = -13 + 5 = -8$
$a_3 = -8 + 5 = -3$

∴ The first three terms are $\mathbf{-13, -8, -3}$.

The salaries form an AP: $5000, 5200, 5400, \ldots$

Here $a = 5000$, $d = 200$ and $a_n = 7000$.

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

$$7000 = 5000 + (n-1) \times 200$$

$$\Rightarrow 200(n-1) = 2000$$

$$\Rightarrow n - 1 = 10$$

$$\Rightarrow n = 11$$

The 1st year is 1995, so the 11th year:
Year = $1995 + 10 = 2005$

∴ His income reached ₹7000 in the 11th year, that is, in 2005.

The weekly savings form an AP: $5, 6.75, 8.50, \ldots$

Here $a = 5$, $d = 1.75$ and $a_n = 20.75$.

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

$$20.75 = 5 + (n-1)(1.75)$$

$$\Rightarrow 1.75(n-1) = 15.75$$

$$\Rightarrow n - 1 = 9$$

$$\Rightarrow n = 10$$

∴ $n = \mathbf{10}$, so in the 10th week her savings become ₹20.75.