# Ex.5.3 Q3 Arithmetic Progressions Solution - NCERT Maths Class 10

## Question

In an AP

(i) Given \(a = 5, d = 3, {a_n} = 50\), find \(n\) and \({S_n}\).

(ii) Given \(a = 7, {a_{13}} = 35\), find \(d\) and \({S_{13}}\).

(iii) Given \({a_{12}} = 37, d = 3\), find \(a\) and \({S_{12}}\).

(iv) Given \({a_3} = 15, {S_{10}} = 125\), find \(d\) and \({a_{10}}\).

(v) Given, find \(a \) and \({a_9}\).

(vi) Given \(a = 2, d = 8, {S_n} = 90\), find \(n\) and \({a_n}\). \(d = 5, {S_9} = 75\)

(vii) Given \(a = 8, {a_n} = 62, {S_n} = 210\), find \(n\) and \(d\).

(viii) Given \({a_n} = 4, d = 2, {S_n} = - 14\), find \(n\) and \(a\).

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

(x) Given \(l = 28, S = 144\) and there are total \(9\) terms. Find \(a\).

## Text Solution

(i) Given \(a = 5, d = 3, {a_n} = 50\), find \(n\) and \({S_n}\).

**What is Known?**

\(a = 5, d = 3, {a_n} = 50\)

**What is Unknown?**

\(n\) and \({S_n}\).

**Reasoning:**

Sum of the first \(n\) terms of an AP is Given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\) or \({S_n} = \frac{n}{2}\left[ {a + l} \right]\), and \(n^\rm{th}\) term of an AP is \(\,{a_n} = a + \left( {n - 1} \right)d\)

Where \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms and \(l\) is the last term.

**Steps:**

Given,

- First term, \(a = - 5\)
- Common difference, \(d = 3\)
- nth term, \(l = {a_n} = 50\)

As \(\,{a_n} = a + \left( {n - 1} \right)d\)

\[\begin{align}50&=5+\left( n-1 \right)3 \\45&=\left( n-1 \right)3 \\15&=n-1 \\n&=16 \\\end{align}\]

\[\begin{align}{S_n} &= \frac{n}{2}\left[ {a + l} \right]\\{S_{16}} &= \frac{{16}}{2}\left[ {5 + 50} \right]\\& = 8 \times 55\\ &= 440\end{align}\]

(ii) Given \(a = 7, {a_{13}} = 35\), find \(d\) and \({S_{13}}\).

**What is Known?**

\(a = 7, {a_{13}} = 35\),

**What is Unknown?**

\(d\) and \({S_{13}}\).

**Reasoning:**

Sum of the first \(n\) terms of an AP is Given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\) or \({S_n} = \frac{n}{2}\left[ {a + l} \right]\), and \(nth\) term of an AP is \(\,{a_n} = a + \left( {n - 1} \right)d\)

Where \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms and \(l\) is the last term.

**Steps:**

**Given,**

- First term, \(a = 7\)
- \(13^\rm{th}\) term, \(l = {a_{13}} = 35\)

As \(\,{a_n} = a + \left( {n - 1} \right)d\)

\[\begin{align}{{a}_{13}}&=a+\left( 13-1 \right)d \\35&=7+12d \\35-7&=12d \\d&=\frac{28}{12} \\ d& =\frac{7}{3} \\\end{align}\]

\[\begin{align}{S_n} &= \frac{n}{2}\left[ {a + l} \right]\\{S_{13}} &= \frac{{13}}{2}\left[ {7 + 35} \right]\\ &= \frac{{13}}{2} \times 42\\ &= 13 \times 21\\ &= 273\end{align}\]

(iii) Given \({a_{12}} = 37, d = 3\), find \(a\) and \({S_{12}}\).

**What is Known?**

\({a_{12}} = 37, d = 3\)

**What is Unknown?**

\(a\) and\({S_{12}}.\)

**Reasoning:**

Sum of the first \(n\) terms of an AP is Given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\) or \({S_n} = \frac{n}{2}\left[ {a + l} \right]\), and term of an AP is \(\,{a_n} = a + \left( {n - 1} \right)d\)

Where \(a\) is the first term, is the common difference and \(n\)is the number of terms and \(l\) is the last term.

**Steps:**

**Given,**

- \(12\)th term, \({a_{12}} = 37\)
- Common Difference, \(d = 3\)

As \(\,{a_n} = a + \left( {n - 1} \right)d\)

\[\begin{align}\,{a_{12}} &= a + \left( {12 - 1} \right)3\\37 &= a + 33\\a &= 4\end{align}\]

\[\begin{align}{S_n}& = \frac{n}{2}\left[ {a + l} \right]\\{S_{12}} &= \frac{{12}}{2}\left[ {4 + 37} \right]\\{S_{12}}& = 6 \times 41\\{S_{12}} &= 246\end{align}\]

(iv) Given \({a_3} = 15, {S_{10}} = 125\), find \(d\) and \({a_{10}}.\)

**What is Known?**

\({a_3} = 15, {S_{10}} = 125\)

**What is Unknown?**

\(d\) and \({a_{10}}.\)

**Reasoning:**

Sum of the first \(n\) terms of an AP is Given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\), and \(n^\rm{th}\) term of an AP is \(\,{a_n} = a + \left( {n - 1} \right)d\)

Where \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms.

**Steps:**

**Given,**

- 3rd term, \({a_3} = 15\)
- Sum up to ten terms, \({S_{10}} = 125\)

As \(\,{a_n} = a + \left( {n - 1} \right)d\)

\[\begin{align}{a_3} &= a + \left( {3 - 1} \right)d\\15 &= a + 2d \qquad \qquad \dots\text{Equation (i )}

\end{align}\]

\[\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\

{S_{10}} &= \frac{{10}}{2}\left[ {2a + \left( {10 - 1} \right)d} \right]\\125 &= 5\left[ {2a + 9d} \right]\\

25 &= 2a + 9d \qquad \qquad \dots\text{Equation (ii)}\end{align}\]

On multiplying equation (i) by \(2,\) we obtain

\[30 = 2a + 4d \qquad \qquad \dots \text{Equation (iii)}\]

On subtracting equation (iii) from Equation (ii), we obtain

\[\begin{align} - 5 &= 5d\\d &= - 1\end{align}\]

From equation (i),

\[\begin{align}15 &= a + 2\left( { - 1} \right)\\15&= a-2\\a& = 17\end{align}\]

\[\begin{align}{a_{10}}& = a + \left( {10 - 1} \right)d\\{a_{10}}& = 17 + 9( - 1)\\{a_{10}} &= 17 - 9\\{a_{10}} &= 8\end{align}\]

(v) Given \(d = 5, {S_9} = 75\), find \(a\) and \({a_9}.\)

**What is Known?**

\(d = 5, {S_9} = 75\)

**What is Unknown?**

\(a\) and \({a_9}.\)

**Reasoning:**

Sum of the first \(n\) terms of an AP is Given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\), and \(n\rm{th}\) term of an AP is \(\,{a_n} = a + \left( {n - 1} \right)d\)

Where \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms.

**Steps:**

**Given,**

- Common difference, \(d = 5\)
- Sum up to nine terms, \({S_9} = 75\)

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

\[\begin{align}{S_9} &= \frac{9}{2}\left[ {2a + \left( {9 - 1} \right)5} \right]\\75 &= \frac{9}{2}\left( {2a + 40} \right)\\25 &= 3\left( {a + 20} \right)\\25 &= 3a + 60\\a &= \frac{{ - 35}}{3}\end{align}\]

We know that \(n\rm{th}\) term of the AP series, \({a_n} = a + \left( {n - 1} \right)d\)

\[\begin{align}{a_9} &= a + \left( {9 - 1} \right) \times 5\\ &= \frac{{ - 35}}{3} + 8 \times 5\\ &= \frac{{ - 35}}{3} + 40\\ &= \frac{{ - 35 + 120}}{3}\\ &= \frac{{85}}{3}\end{align}\]

(vi) Given \(a = 2, d = 8, {S_n} = 90\), find \(n\) and \({a_n}.\)

**What is Known?**

\(a = 2, d = 8, {S_n} = 90\)

**What is Unknown?**

\(n\) and \({a_n}.\)

**Reasoning:**

Sum of the first \(n\) terms of an AP is Given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\) or \({S_n} = \frac{n}{2}\left[ {a + l} \right]\), and \(n\rm{th}\) term of an AP is\(\,{a_n} = a + \left( {n - 1} \right)d\)

Where \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms and \(l\) is the last term.

**Steps:**

**Given,**

- First term, \(a = 2\)
- Common difference, \(d = 5\)
- Sum up to nth terms, \({S_n} = 90\)

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

\[\begin{align}90 = \frac{n}{2}\left[ {4 + \left( {n - 1} \right)8} \right]\\90{\rm{ }} = n{\rm{ }}[2 + (n - 1)4]\\

90 = n\left[ {2 + 4n - 4} \right]\\90 = n\left[ {4n - 2} \right]\\90 = 4{n^2} - 2n\\4{n^2} - 2n - 90 = 0\\4{n^2} - 20n + 18n - 90 = 0\\4n\left( {n - 5} \right) + 18\left( {n - 5} \right) = 0\\\left( {n - 5} \right)\left( {4n + 18} \right) = 0\end{align}\]

Either \((n - 5) = 0\) or \((4n + 18) = 0\)

\(n = 5\,\) or \(n = \frac{{ - 9}}{2}\)

However, \(n\) can neither be negative nor fractional.

Therefore, \(n = 5\,\)

\[\begin{align}{a_n} &= a + \left( {n - 1} \right)d\\{a_5} &= 2 + \left( {5 - 1} \right)8\\{a_5} &= 2 + 4 \times 8\\{a_5} &= 2 + 32\\{a_5} &= {\rm{ }}34\end{align}\]

(vii) Given \(a = 8, {a_n} = 62, {S_n} = 210\), find and \(d\).

**What is Known?**

\(a = 8, {a_n} = 62, {S_n} = 210\)

**What is Unknown?**

\(n\) and \(d.\)

**Reasoning:**

Sum of the first \(n\) terms of an AP is Given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\) or \({S_n} = \frac{n}{2}\left[ {a + l} \right]\), and \(nth\) term of an AP is \(\,{a_n} = a + \left( {n - 1} \right)d\)

Where \(a\) is the first term, \(d\) is the common difference and \(n\)is the number of terms and \(l\) is the last term.

**Steps:**

**Given,**

- First term, \(a = 8\)
- \(n^\rm{th}\) term, \(l = {a_n} = 62\)
- Sum up to nth terms, \({S_n} = 210\)

As \({S_n} = \frac{n}{2}\left[ {a + l} \right]\)

\[\begin{align}210& = \frac{n}{2}\left[ {8 + 62} \right]\\210 &= \frac{n}{2} \times 70\\n &= 6\end{align}\]

We know that \(n^\rm{th}\) term of the AP series, \({a_n} = a + \left( {n - 1} \right)d\)

\[\begin{align}62 &= 8 + \left( {6 - 1} \right)d\\62 - 8 &= 5d\\54 &= 5d\\d &= \frac{{54}}{5}\end{align}\]

(viii) Given \({a_n} = 4,{\rm{ }}d = 2,{\rm{ }}{S_n} = - {\rm{ }}14,\) find \(n\) and \(a\).

**What is Known?**

\({a_n} = 4, d = 2, {S_n} = - 14\)

**What is Unknown?**

\(n\) and \(a\)

**Reasoning:**

Sum of the first \(n\) terms of an AP is Given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\) or \({S_n} = \frac{n}{2}\left[ {a + l} \right]\), and \(nth\) term of an AP is \(\,{a_n} = a + \left( {n - 1} \right)d\)

**Steps:**

**Given,**

- Common difference, \(d = 2\)
- \(n^\rm{th}\) term, \(l = {a_n} = 4\)
- Sum up to nth terms, \({S_n} = - 14\)

We know that nth term of AP series,\(\,{a_n} = a + \left( {n - 1} \right)d\)

\[\begin{align}4 &= a + (n - 1)2\\4 &= a + 2n - 2\\a &= 6 - 2n \qquad \dots(1)\end{align}\]

\[\begin{align}{S_n} &= \frac{n}{2}\left[ {a + l} \right]\\ - 14 &= \frac{n}{2}\left[ {6 - 2n + 4} \right].............[\text{from(1)}]\\ - 14& = n\left( {5 - n} \right)\\ - 14 &= 5n - {n^2}\\{n^2} - 5n - 14 &= 0\\{n^2} - 7n + 2n - 14 &= 0\\n\left( {n - 7} \right) + 2\left( {n - 7} \right) &= 0\\\left( {n - 7} \right)\left( {n + 2} \right) &= 0\end{align}\]

Either \(n - 7 = 0\) or \(n + 2 = 0\)

\(n=7\) or \(n = - 2\)

However, \(n\) can neither be negative nor fractional.

Therefore, \(n=7\)

From equation (1), we obtain

\[\begin{align}a&= 6 - 2n\\a& = 6 - 2 \times 7\\a& = 6-14\\a& = - 8\end{align}\]

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

**What is Known?**

\(a = 3, n = 8, S = 192\)

**What is Unknown?**

\(d\)

**Reasoning:**

Sum of the first \(n\) terms of an AP is Given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\)

Where \(a\) is the first term, \(d\) is the common difference and \(n\)is the number of terms.

**Steps:**

**Given,**

- First term, \(a = 3\)
- Number of terms, \(n = 8\)
- Sum up to nth terms, \({S_n} = 192\)

\[\begin{align}{S_n}& = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\192 &= \frac{8}{2}\left[ {2 \times 3 + \left( {8 - 1} \right)d} \right]\\192 &= 4\left[ {6 + 7d} \right]\\48 &= 6 + 7d\\42 &= 7d\\d &= 6\end{align}\]

(x) Given \(l = 28, S = 144\) and there are total \(9\) terms. Find \(a\).

**What is Known?**

\(l = 28, S = 144, n = 9\)

**What is Unknown?**

\(a\)

**Reasoning:**

Sum of the first \(n\) terms of an AP is Given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\) or \({S_n} = \frac{n}{2}\left[ {a + l} \right]\)

Where \(a\) is the first term, \(d\) is the common difference and \(n\)is the number of terms and \(l\) is the last term.

**Steps:**

**Given,**

- Last term, \(l = {a_n} = 28\)
- Number of terms, \(n = 9\)
- Sum up to nth terms, \({S_n} = 192\)

\[\begin{align}{S_n} &= \frac{n}{2}\left( {a + l} \right)\\144 &= \frac{9}{2}\left( {a + 28} \right)\\32 &= a + 28\\a &= 4\end{align}\]