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

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## 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 $$a$$.

Video Solution
Arithmetic Progressions
Ex 5.3 | Question 3

## 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{(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 \quad \dots\text{(ii)}\end{align}

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

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

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

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,

• 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]\dots[\text{from(1)}]\\ - 14& = n\left( {5 - n} \right)\\ - 14 &= 5n - {n^2}\\{n^2} \!-\! 5n \!-\! 14\! &= 0\\\begin{bmatrix}{n^2}\! - \!7n + \\2n - 14 \end{bmatrix}&= 0\\\begin{bmatrix}\left( {n - 7} \right) + \\2\left( {n - 7} \right) \end{bmatrix}&= 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}

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