# NCERT Solutions For Class 11 Maths Chapter 9 Exercise 9.2

## Chapter 9 Ex.9.2 Question 1

Find the sum of odd integers from $$1$$ to $$2001$$.

### Solution

The odd integers from $$1$$ to $$2001$$ are $$1,3,5,7,9, \ldots ,1999,2001$$.

This sequence forms an A.P.

Here, first term, $$a = 1$$

Common difference, $$d = 2$$

Last term $$l = {a_n} = 2001$$

Therefore,

\begin{align}{a_n} &= a + \left( {n - 1} \right)d\\1 + \left( {n - 1} \right)\left( 2 \right) &= 2001\\1 + 2n - 2 &= 2001\\2n& = 2001 + 1\\n &= \frac{{2002}}{2}\\n &= 1001\end{align}

Now,

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{1001}} &= \frac{{1001}}{2}\left[ {2 \times 1 + \left( {1001 - 1} \right) \times 2} \right]\\&= \frac{{1001}}{2}\left[ {2 + 1000 \times 2} \right]\\&= \frac{{1001}}{2} \times 2002\\&= 1001 \times 1001\\&= 1002001\end{align}

Thus, the sum of odd numbers from $$1$$ to $$2001$$ is $$1002001$$.

## Chapter 9 Ex.9.2 Question 2

Find the sum of all-natural numbers lying between 100 and 1000, which are multiples of 5.

### Solution

The natural numbers lying between $$100$$ and $$1000$$, which are multiples of $$5$$ are

$$105,110,115, \ldots ,995$$

Here, first term, $$a = 105$$

Common difference, $$d = 5$$

Last term $$l = {a_n} = 995$$

Therefore,

\begin{align}{a_n}&= a + \left( {n - 1} \right)d\\105 + \left( {n - 1} \right)5 &= 995\\\left( {n - 1} \right)5& = 995 - 105\\\left( {n - 1} \right) &= \frac{{890}}{5}\\n &= 178 + 1\\n &= 179\end{align}

Now,

\begin{align}{c}{S_n} &= \frac{{179}}{2}\left[ {2\left( {105} \right) + \left( {179 - 1} \right)\left( 5 \right)} \right]\\&= \frac{{179}}{2}\left[ {2\left( {105} \right) + \left( {178} \right)\left( 5 \right)} \right]\\&= 179\left[ {105 + 89\left( 5 \right)} \right]\\&= 179\left[ {105 + 445} \right]\\&= 179 \times 550\\&= 98450\end{align}

Thus, the sum of all-natural numbers lying between $$100$$ and $$1000$$, which are multiples of 5 is $$98450$$ .

## Chapter 9 Ex.9.2 Question 3

In an A.P, the first term is $$2$$ and the sum of the first five terms is one-fourth of the next five terms. Show that $${20^{th}}$$ term is $$- 112$$.

### Solution

It is given that the first term of the A.P. $$a = 2$$

Let $$d$$ be the common difference of the A.P.

Hence, the A.P is $$2,2 + d,2 + 2d,2 + 3d \ldots$$

We know that, $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$

Sum of the first five terms

\begin{align}{S_5}&= \frac{5}{2}\left[ {2 \times 2 + \left( {5 - 1} \right)d} \right]\\&= \frac{5}{2}\left[ {4 + 4d} \right]\\&= 10 + 10d\end{align}

Sum of the first ten terms

\begin{align}{S_{10}}&= \frac{{10}}{2}\left[ {2 \times 2 + \left( {10 - 1} \right)d} \right]\\&= 5\left[ {4 + 9d} \right]\\&= 20 + 45d\end{align}

Hence,

Sum of the next five terms

\begin{align}{S_{10}} - {S_5}&= \left( {20 + 45d} \right) - \left( {10 + 10d} \right)\\&= 20 + 45d - 10 - 10d\\&= 10 + 35d\end{align}

According to the given condition,

\begin{align}&\Rightarrow 10 + 10d = \frac{1}{4}\left( {10 + 35d} \right)\\&\Rightarrow 40 + 40d = 10 + 35d\\&\Rightarrow 40d - 35d = 10 - 40\\&\Rightarrow 5d = - 30\\&\Rightarrow d = - 6\end{align}

Therefore,

\begin{align}{a_{20}} &= a + \left( {20 - 1} \right)d\\&= 2 + \left( {19} \right)\left( { - 6} \right)\\&= 2 - 114\\&= - 112\end{align}

Thus, the $${20^{th}}$$ term of the A.P is $$- 112$$.

## Chapter 9 Ex.9.2 Question 4

How many terms of the A.P. $$- 6,\frac{{ - 11}}{2}, - 5, \ldots$$are needed to give the sum $$- 25$$?

### Solution

It is given that $$- 6,\frac{{ - 11}}{2}, - 5, \ldots$$ are in A.P.

Let the sum of $$n$$ terms of the given A.P be $${S_n} = - 25$$.

Here, first term $$a = - 6$$

Common difference $$d = \frac{{ - 11}}{2} + 6 = \frac{{ - 11 + 12}}{2} = \frac{1}{2}$$

We know that, $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$

Therefore,

\begin{align}&\Rightarrow - 25 = \frac{n}{2}\left[ {2 \times \left( { - 6} \right) + \left( {n - 1} \right)\left( {\frac{1}{2}} \right)} \right]\\&\Rightarrow - 50 = n\left[ { - 12 + \frac{n}{2} - \frac{1}{2}} \right]\\&\Rightarrow - 50 = n\left[ {\frac{n}{2} - \frac{{25}}{2}} \right]\\&\Rightarrow - 100 = n\left[ {n - 25} \right]\\&\Rightarrow {n^2} - 25n + 100 = 0\\&\Rightarrow {n^2} - 5n - 20n + 100 = 0\\&\Rightarrow n\left( {n - 5} \right) - 20\left( {n - 5} \right) = 0\\&\Rightarrow \left( {n - 5} \right)\left( {n - 20} \right) = 0\\&\Rightarrow n = 20,5\end{align}

## Chapter 9 Ex.9.2 Question 5

In an A.P, if the $${p^{th}}$$ term is $$\frac{1}{q}$$ and $${q^{th}}$$ term is $$\frac{1}{p}$$, prove that the sum of first $$pq$$ terms is $$\frac{1}{2}\left( {pq + 1} \right)$$, where$$p \ne q$$.

### Solution

It is known that general term of an A.P is $${a_n} = a + \left( {n - 1} \right)d$$

Therefore,

\begin{align}{p^{th}}term &= {a_p} = a + \left( {p - 1} \right)d = \frac{1}{q}\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\{q^{th}}term&= {a_q} = a + \left( {q - 1} \right)d = \frac{1}{p}\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Subtracting $$\left( 2 \right)$$from $$\left( 1 \right)$$, we obtain

\begin{align}&\Rightarrow \left( {p - 1} \right)d - \left( {q - 1} \right)d = \frac{1}{q} - \frac{1}{p}\\&\Rightarrow \left( {p - 1 - q + 1} \right)d = \frac{{p - q}}{{pq}}\\&\Rightarrow \left( {p - q} \right)d = \frac{{p - q}}{{pq}}\\&\Rightarrow \left( {p - q} \right)d = \frac{{p - q}}{{pq}}\\&\Rightarrow d = \frac{1}{{pq}}\end{align}

Putting the value of $$d$$ in $$\left( 1 \right)$$, we obtain

\begin{align}& \Rightarrow a + \left( {p - 1} \right)\frac{1}{{pq}} = \frac{1}{q}\\ &\Rightarrow a = \frac{1}{q} - \frac{1}{q} + \frac{1}{{pq}} = \frac{1}{{pq}}\end{align}

Therefore,

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

Thus, the sum of first $$pq$$ terms of the A.P is $$\frac{1}{2}\left( pq+1 \right)$$.

## Chapter 9 Ex.9.2 Question 6

If the sum of a certain number of terms of the A.P $$25,22,19, \ldots$$is $${\rm{116}}$$. Find the last term.

### Solution

It is given that $$25,22,19, \ldots$$ are in A.P.

Let the sum of $$n$$ terms of an A.P be $${\rm{116}}$$.

Here, first term $$a = 25$$

Common di$$n$$fference $$d = 22 - 25 = - 3$$

We know that, $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$

Therefore,

\begin{align}&\Rightarrow 116= \frac{n}{2}\left[ {2\left( {25} \right) + \left( {n - 1} \right)\left( { - 3} \right)} \right]\\&\Rightarrow 116= \frac{n}{2}\left[ {50 - 3n + 3} \right]\\&\Rightarrow 232= n\left( {53 - 3n} \right) = 53n - 3{n^2}\\&\Rightarrow 3{n^2} - 53n + 232 &= 0\\&\Rightarrow 3{n^2} - 24n - 29n + 232 = 0\\&\Rightarrow 3n\left( {n - 8} \right) - 29\left( {n - 8} \right) = 0\\&\Rightarrow \left( {n - 8} \right)\left( {3n - 29} \right) = 0\\&\Rightarrow n = 8,\frac{{29}}{3}\end{align}

However, cannot be equal to $$\frac{{29}}{3}$$

Hence, $$n = 8$$

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

\begin{align}{a_8}& = 25 + \left( {8 - 1} \right)\left( { - 3} \right)\\&= 25 + \left( 7 \right)\left( { - 3} \right)\\&= 25 - 21\\&= 4\end{align}

Thus, the last term of the A.P is $$4$$.

## Chapter 9 Ex.9.2 Question 7

Find the sum $$n$$ to terms of the A.P, whose $${{k}^{th}}$$ term is $$5k+1$$.

### Solution

It is given that $${{k}^{th}}$$ term of the A.P is $$5k+1$$.

i.e., $$n{{a}_{k}}=\left( 5k+1 \right)$$

It is known that general term of an A.P is $${a_n} = a + \left( {n - 1} \right)d$$

\begin{align}{a_k} &= a + \left( {k - 1} \right)d\\5k + 1 &= a + \left( {k - 1} \right)d\\5k + 1 &= kd + \left( {a - d} \right)\end{align}

Comparing the coefficients of $$k$$, we obtain $$d = 5$$ and

\begin{align}\Rightarrow a - d&= 1\\\Rightarrow a - 5&= 1\\\Rightarrow a&= 6\end{align}

Therefore,

\begin{align}{S_n}&= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\&= \frac{n}{2}\left[ {2\left( 6 \right) + \left( {n - 1} \right)\left( 5 \right)} \right]\\&= \frac{n}{2}\left[ {12 + 5n - 5} \right]\\&= \frac{n}{2}\left[ {5n + 7} \right]\end{align}

## Chapter 9 Ex.9.2 Question 8

If the sum of $$n$$terms of an A.P is $$\left( {pn + q{n^2}} \right)$$, where $$p$$ and $$q$$ are constants, find the common difference.

### Solution

We know that $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$

According to the question,

\begin{align}\frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]&= pn + q{n^2}\\\frac{n}{2}\left[ {2a + nd - d} \right]&= pn + q{n^2}\\an + \frac{1}{2}d{n^2} - \frac{1}{2}dn&= pn + q{n^2}\\\left( {a - \frac{1}{2}d} \right)n + \frac{1}{2}d{n^2}&= pn + q{n^2}\end{align}

Comparing the coefficients of $${n^2}$$ on both sides, we obtain

\begin{align}\Rightarrow \frac{1}{2}d = q\\\Rightarrow d = 2q\end{align}

Thus, the common difference is $$2q$$.

## Chapter 9 Ex.9.2 Question 9

The sums of $$n$$ terms of two arithmetic progressions are in the ratio $$5n + 4:9n + 6$$. Find the ratio of their $${18^{th}}$$ terms.

### Solution

Let $${a_1},\;{a_2}$$ and $${d_1},\;{d_2}$$ be the first terms and common differences of two arithmetic progressions respectively.

According to the given condition,

$$\frac{{{\text{Sum of n terms of first A}}.{\text{P}}}}{{{\text{Sum of n terms of second A}}.{\rm{P}}}} = \frac{{5n + 4}}{{9n + 6}}$$

\begin{align}&\Rightarrow \frac{{\frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right){d_1}} \right]}}{{\frac{n}{2}\left[ {2{a_2} + \left( {n - 1} \right){d_2}} \right]}} = \frac{{5n + 4}}{{9n + 6}}\\&\Rightarrow \frac{{2{a_1} + \left( {n - 1} \right){d_1}}}{{2{a_2} + \left( {n - 1} \right){d_2}}} = \frac{{5n + 4}}{{9n + 6}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Substituting $$n = 35$$ in $$\left( 1 \right)$$, we obtain

\begin{align}&\Rightarrow \frac{{2{a_1} + \left( {34} \right){d_1}}}{{2{a_2} + \left( {34} \right){d_2}}} = \frac{{5\left( {35} \right) + 4}}{{9\left( {35} \right) + 6}}\\&\Rightarrow \frac{{{a_1} + 17{d_1}}}{{{a_2} + 17{d_2}}} = \frac{{179}}{{321}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Now,

$$\frac{{{\rm{1}}{{\text{8}}^{th}}{\text{term of first A}}.{\text{P}}}}{{{\text{1}}{{\text{8}}^{th}}{\text{term of second A}}.{\rm{P}}}} = \frac{{{a_1} + 17{d_1}}}{{{a_2} + 17{d_2}}}\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)$$

From ($$2$$) and ($$3$$), we obtain

$$\frac{{{\text{1}}{{\rm{8}}^{th}}{\text{term of first A}}.{\text{P}}}}{{{\rm{1}}{{\text{8}}^{th}}{\text{term of second A}}.{\text{P}}}} = \frac{{179}}{{321}}$$

Thus, the ratio of their $${18^{th}}$$ terms is $$179:321$$.

## Chapter 9 Ex.9.2 Question 10

If the sum of first $$p$$ terms of an A.P is equal to the sum of first $$q$$ terms, then find the sum of first $$\left( {p + q} \right)$$ terms.

### Solution

Let $$a$$ and $$d$$ be the first term and the common difference of the A.P. respectively.

We know that $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$

Hence,

\begin{align}{S_p} &= \frac{p}{2}\left[ {2a + \left( {p - 1} \right)d} \right]\\{S_q} &= \frac{q}{2}\left[ {2a + \left( {q - 1} \right)d} \right]\end{align}

According to the question,

\begin{align}&\Rightarrow \frac{p}{2}\left[ {2a + \left( {p - 1} \right)d} \right] = \frac{q}{2}\left[ {2a + \left( {q - 1} \right)d} \right]\\&\Rightarrow p\left[ {2a + \left( {p - 1} \right)d} \right] = q\left[ {2a + \left( {q - 1} \right)d} \right]\\&\Rightarrow 2ap + pd\left( {p - 1} \right) = 2aq + qd\left( {q - 1} \right)\\&\Rightarrow 2a\left( {p - q} \right) + d\left[ {p\left( {p - 1} \right) - q\left( {q - 1} \right)} \right] = 0\\&\Rightarrow 2a\left( {p - q} \right) + d\left[ {{p^2} - p - {q^2} + q} \right] = 0\\&\Rightarrow 2a\left( {p - q} \right) + d\left[ {\left( {p - q} \right)\left( {p + q} \right) - \left( {p - q} \right)} \right] = 0\\&\Rightarrow 2a\left( {p - q} \right) + d\left[ {\left( {p - q} \right)\left( {p + q - 1} \right)} \right] = 0\\&\Rightarrow 2a + d\left( {p + q - 1} \right) = 0\\&\Rightarrow d = \frac{{ - 2a}}{{p + q - 1}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Now,

\begin{align}{S_{p + q}}&= \frac{{p + q}}{2}\left[ {2a + \left( {p + q - 1} \right)\left( {\frac{{ - 2a}}{{p + q - 1}}} \right)} \right]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {from\left( 1 \right)} \right]\\&= \frac{{p + q}}{2}\left[ {2a - 2a} \right]\\&= 0\end{align}

Thus, the sum of the first $$\left( {p + q} \right)$$ terms of the A.P is 0.

## Chapter 9 Ex.9.2 Question 11

Sum of the first $$p,\;q$$ and $$r$$ terms of an A.P., are $$a,\;b$$ and $$c$$ respectively. Prove that $$\frac{a}{p}\left( {q - r} \right) + \frac{b}{q}\left( {r - p} \right) + \frac{c}{r}\left( {p - q} \right) = 0$$

### Solution

Let $${a_1}$$ and $$d$$ be the first term and the common difference of the given arithmetic progression respectively.

We know that $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$

According to the given information,

\begin{align}{S_p}&= \frac{p}{2}\left[ {2{a_1} + \left( {p - 1} \right)d} \right]\\a& = \frac{p}{2}\left[ {2{a_1} + \left( {p - 1} \right)d} \right]\\&\Rightarrow 2{a_1} + \left( {p - 1} \right)d = \frac{{2a}}{p}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\\\{S_q} &= \frac{q}{2}\left[ {2{a_1} + \left( {q - 1} \right)d} \right]\\b &= \frac{q}{2}\left[ {2{a_1} + \left( {q - 1} \right)d} \right]\\&\Rightarrow 2{a_1} + \left( {q - 1} \right)d = \frac{{2b}}{q}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\\\{S_r} &= \frac{r}{2}\left[ {2{a_1} + \left( {r - 1} \right)d} \right]\\c &= \frac{r}{2}\left[ {2{a_1} + \left( {r - 1} \right)d} \right]\\&\Rightarrow 2{a_1} + \left( {r - 1} \right)d = \frac{{2c}}{r}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}

Subtracting $$\left( 2 \right)$$ from $$\left( 1 \right)$$

\begin{align}&\Rightarrow \left( {p - 1} \right)d - \left( {q - 1} \right)d = \frac{{2a}}{p} - \frac{{2b}}{q}\\&\Rightarrow d\left( {p - 1 - q + 1} \right) = \frac{{2aq - 2bp}}{{pq}}\\&\Rightarrow d\left( {p - q} \right) = \frac{{2aq - 2bp}}{{pq}}\\&\Rightarrow d = \frac{{2\left( {aq - bp} \right)}}{{pq\left( {p - q} \right)}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 4 \right)\end{align}

Subtracting $$\left( 3 \right)$$from $$\left( 2 \right)$$

\begin{align}&\Rightarrow \left( {q - 1} \right)d - \left( {r - 1} \right)d = \frac{{2b}}{q} - \frac{{2c}}{r}\\&\Rightarrow d\left( {q - 1 - r + 1} \right) = \frac{{2b}}{q} - \frac{{2c}}{r}\\&\Rightarrow d\left( {q - r} \right) = \frac{{2br - 2cq}}{{qr}}\\&\Rightarrow d = \frac{{2\left( {br - qc} \right)}}{{qr\left( {q - r} \right)}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 5 \right)\end{align}

Equating both the values of $$d$$ obtained in $$\left( 4 \right)$$and $$\left( 5 \right)$$, we obtain

\begin{align}&\Rightarrow \frac{{aq - bp}}{{pq\left( {p - q} \right)}} = \frac{{br - qc}}{{qr\left( {q - r} \right)}}\\&\Rightarrow qr\left( {q - r} \right)\left( {aq - bp} \right) = pq\left( {p - q} \right)\left( {br - qc} \right)\\&\Rightarrow r\left( {aq - bp} \right)\left( {q - r} \right) = p\left( {br - qc} \right)\left( {p - q} \right)\\&\Rightarrow \left( {aqr - bpr} \right)\left( {q - r} \right) = \left( {bpr - pqc} \right)\left( {p - q} \right)\end{align}

Dividing both sides by $$pqr$$, we obtain

\begin{align}&\Rightarrow \left( {\frac{a}{p} - \frac{b}{q}} \right)\left( {q - r} \right) = \left( {\frac{b}{q} - \frac{c}{r}} \right)\left( {p - q} \right)\\&\Rightarrow \frac{a}{p}\left( {q - r} \right) - \frac{b}{q}\left( {q - r} \right) = \frac{b}{q}\left( {p - q} \right) - \frac{c}{r}\left( {p - q} \right)\\&\Rightarrow \frac{a}{p}\left( {q - r} \right) - \frac{b}{q}\left( {q - r + p - q} \right) + \frac{c}{r}\left( {p - q} \right) = 0\\&\Rightarrow \frac{a}{p}\left( {q - r} \right) + \frac{b}{q}\left( {r - p} \right) + \frac{c}{r}\left( {p - q} \right) = 0\end{align}

Hence, proved.

## Chapter 9 Ex.9.2 Question 12

The ratio of the sum of $$m$$ and terms of an A.P is $${m^2}:{n^2}$$. Show that the ratio of $${m^{th}}$$ and $${n^{th}}$$ term is $$\left( {2m - 1} \right):\left( {2n - 1} \right)$$.

### Solution

Let $$a$$ and $$d$$ be the first term and the common difference of the given arithmetic progression respectively.

According to the given condition,

\begin{align} &\Rightarrow \frac{\text{Sum of m terms}}{\text{Sum of n terms}}=\frac{{{m}^{2}}}{{{n}^{2}}} \\ &\Rightarrow \frac{\frac{m}{2}\left[ 2a+\left( m-1 \right)d \right]}{\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]}=\frac{{{m}^{2}}}{{{n}^{2}}} \\ & \Rightarrow \frac{2a+\left( m-1 \right)d}{2a+\left( n-1 \right)d}=\frac{m}{n}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 1 \right) \\ \end{align}

Putting $$m = 2m - 1$$ and $$n = 2n - 1$$, we obtain

\begin{align}&\frac{{2a + \left( {2m - 2} \right)d}}{{2a + \left( {2n - 2} \right)d}} = \frac{{2m - 1}}{{2n - 1}}\\&\Rightarrow \frac{{a + \left( {m - 1} \right)d}}{{a + \left( {n - 1} \right)d}} = \frac{{2m - 1}}{{2n - 1}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\&\frac{{{m^{th}}{\rm{ term of A}}.{\rm{P}}}}{{{n^{th}}{\rm{ term of A}}.{\rm{P}}}} = \frac{{a + \left( {m - 1} \right)d}}{{a + \left( {n - 1} \right)d}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}

From $$\left( 2 \right)$$and $$\left( 3 \right)$$, we obtain$$\frac{{{{\rm{m}}^{th}}{\rm{ term of A}}.{\rm{P}}}}{{{n^{th}}{\rm{ term of A}}.{\rm{P}}}} = \frac{{2m - 1}}{{2n - 1}}$$

$\frac{{{{\rm{m}}^{th}}{\rm{ term of A}}.{\rm{P}}}}{{{n^{th}}{\rm{ term of A}}.{\rm{P}}}} = \frac{{2m - 1}}{{2n - 1}}$

Hence, proved.

## Chapter 9 Ex.9.2 Question 13

If the sum of terms of an A.P of $$3{n^2} + 5n$$ and its $${m^{th}}$$ term is $$164$$, find the value of $$m$$.

### Solution

Let $$a$$ and $$d$$ be the first term and the common difference of the given arithmetic progression respectively.

$${a_m} = a + \left( {m - 1} \right)d = 164\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)$$

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

Here,

\begin{align}&\Rightarrow \frac{n}{2}\left[ {2a + nd - d} \right] = 3{n^2} + 5n\\&\Rightarrow an + \frac{d}{2}{n^2} - \frac{d}{2}n = 3{n^2} + 5n\\&\Rightarrow \frac{d}{2}{n^2} + \left( {a - \frac{d}{2}} \right)n = 3{n^2} + 5n\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Comparing the coefficient of $${n^2}$$ on both sides in $$\left( 2 \right)$$, we obtain

\begin{align}&\Rightarrow \frac{d}{2} = 3\\&\Rightarrow d = 6\end{align}

Comparing the coefficient of on both sides in $$\left( 2 \right)$$, we obtain

\begin{align} & \Rightarrow a-\frac{d}{2}=5 \\ & \Rightarrow a-3=5 \\ & \Rightarrow a=8 \\ \end{align}

Therefore, from $$\left( 1 \right)$$

\begin{align}&\Rightarrow 8 + \left( {m - 1} \right)6 = 164\\&\Rightarrow \left( {m - 1} \right)6 = 164 - 8\\&\Rightarrow \left( {m - 1} \right) = \frac{{156}}{6}\\&\Rightarrow m = 26 + 1\\&\Rightarrow m = 27\end{align}

Thus, the value of $$m = 27$$.

## Chapter 9 Ex.9.2 Question 14

Insert five numbers between $$8$$ and $$26$$ such that resulting sequence is an A.P.

### Solution

Let $${A_1},{A_2},{A_3},{A_4}$$ and $${A_5}$$ be the five numbers between 8 and 26 such that;

$$8,{A_1},{A_2},{A_3},{A_4},{A_5},26$$ are in A.P.

Here, $$a = 8,\;b = 26,\;n = 7$$

Hence,

\begin{align}\Rightarrow 26 &= 8 + \left( {7 - 1} \right)d\\\Rightarrow 26 &= 8 + \left( 6 \right)d\\\Rightarrow 6d &= 26 - 8\\\Rightarrow 6d &= 18\\\Rightarrow d &= 3\end{align}

Therefore,

\begin{align}{A_1} &= a + d = 8 + 3 = 11\\{A_2}&= a + 2d = 8 + \left( 2 \right)3 = 14\\{A_3} &= a + 3d = 8 + \left( 3 \right)3 = 17\\{A_4}&= a + 4d = 8 + \left( 4 \right)3 = 20\\{A_5}&= a + 5d = 8 + \left( 5 \right)3 = 23\end{align}

Thus, the required five numbers between $$8$$ and $$26$$ are $$11,14,17,20$$ and $$23$$.

## Chapter 9 Ex.9.2 Question 15

If $$\frac{{{a^n} + {b^n}}}{{{a^{n - 1}} + {b^{n - 1}}}}$$ is the A.M between $$a$$ and $$b$$, then find the value of $$n$$.

### Solution

A.M of $$a$$ and $$b = \frac{{a + b}}{2}$$

According to the question,

\begin{align}&\Rightarrow \frac{{a + b}}{2} = \frac{{{a^n} + {b^n}}}{{{a^{n - 1}} + {b^{n - 1}}}}\\&\Rightarrow \left( {a + b} \right)\left( {{a^{n - 1}} + {b^{n - 1}}} \right) = 2\left( {{a^n} + {b^n}} \right)\\&\Rightarrow {a^n} + a{b^{n - 1}} + b{a^{n - 1}} + {b^n} = 2{a^n} + 2{b^n}\\&\Rightarrow a{b^{n - 1}} + b{a^{n - 1}} = {a^n} + {b^n}\\&\Rightarrow a{b^{n - 1}} - {b^n} = {a^n} - b{a^{n - 1}}\\&\Rightarrow {b^{n - 1}}\left( {a - b} \right) = {a^{n - 1}}\left( {a - b} \right)\\&\Rightarrow {b^{n - 1}} = {a^{n - 1}}\\&\Rightarrow {\left( {\frac{a}{b}} \right)^{n - 1}} = 1 = {\left( {\frac{a}{b}} \right)^0}\\&\Rightarrow n - 1 = 0\\&\Rightarrow n = 1\end{align}

## Chapter 9 Ex.9.2 Question 16

Between 1 and 31, $$m$$ numbers have been inserted in such a way that the resulting sequence is an A.P and the ratio of $${7^{th}}$$ and $${\left( {m - 1} \right)^{th}}$$ numbers is $$5:9$$. Find the value of $$m$$.

### Solution

Let $${A_1},{A_2}, \ldots ,{A_m}$$ be numbers such that $$1,{A_1},{A_2}, \ldots ,{A_m},31$$ are in A.P.

Here, $$a = 1,\;b = 31,\;n = m + 2$$

Therefore,

\begin{align}&\Rightarrow 31 = 1 + \left( {m + 2 - 1} \right)d\\&\Rightarrow 30 = \left( {m + 1} \right)d\\&\Rightarrow d = \frac{{30}}{{m + 1}}\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\\end{align}

Hence,

\begin{align}{A_1} &= a + d\\{A_2}&= a + 2d\\{A_3}&= a + 3d\\{A_7}&= a + 7d\\{A_{m - 1}} &= a + \left( {m - 1} \right)d\end{align}

According to the given condition,

\begin{align}&\Rightarrow \frac{{a + 7d}}{{a + \left( {m - 1} \right)d}} = \frac{5}{9}\\&\Rightarrow \frac{{1 + 7\left( {\frac{{30}}{{m + 1}}} \right)}}{{1 + \left( {m - 1} \right)\left( {\frac{{30}}{{m + 1}}} \right)}} = \frac{5}{9}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {from{\rm{ }}\left( 1 \right)} \right]\\&\Rightarrow \frac{{m + 1 + 7\left( {30} \right)}}{{m + 1 + 30\left( {m - 1} \right)}} = \frac{5}{9}\\&\Rightarrow \frac{{m + 1 + 210}}{{m + 1 + 30m - 30}} = \frac{5}{9}\\&\Rightarrow \frac{{m + 211}}{{31m - 29}} = \frac{5}{9}\\&\Rightarrow 9m + 1899 = 155m - 145\\&\Rightarrow 155m - 9m = 1899 + 145\\&\Rightarrow 146m = 2044\\&\Rightarrow m = 14\end{align}

Thus, the value of $$m = 14$$.

## Chapter 9 Ex.9.2 Question 17

A man starts repaying a loan as first instalment of ? $$100$$. If he increases the instalment by ? $$5$$every month, what amount will he repay in the $${30^{th}}$$ instalment?

### Solution

The first installment of the load is $$₹$$ $$100$$.

The second installment of the load is $$₹$$ $$105$$ and so on.

The amount that the man repays every month becomes an A.P.

The A.P is $$100,105,110, \ldots$$

Here, $$a = 100,d = 5$$

\begin{align}{A_{30}}&= a + \left( {30 - 1} \right)d\\&= 100 + 29\left( 5 \right)\\&= 100 + 145\\&= 245\end{align}

Thus, the amount to be paid in the $${30^{th}}$$ instalment is $$₹$$ $$245$$.

## Chapter 9 Ex.9.2 Question 18

The difference between any two consecutive interior angles of a polygon is $$5^\circ$$. If the smallest angle is $$120^\circ$$, find the number of the sides of the polygon.

### Solution

The angles of the polygon will form an A.P with $$d = 5^\circ$$ and $$a = 120^\circ$$.

We know that the sum of all angles of a polygon with $$n$$ sides is $$180^\circ \left( {n - 2} \right)$$

Therefore,

\begin{align}&\Rightarrow {S_n} = 180^\circ \left( {n - 2} \right)\\&\Rightarrow \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] = {180^ \circ }\left( {n - 2} \right)\\&\Rightarrow \frac{n}{2}\left[ {{{240}^ \circ } + \left( {n - 1} \right){5^ \circ }} \right] = {180^ \circ }\left( {n - 2} \right)\\&\Rightarrow n\left[ {240 + \left( {n - 1} \right)5} \right] = 360\left( {n - 2} \right)\\&\Rightarrow 240n + 5{n^2} - 5n = 360n - 720\\&\Rightarrow 5{n^2} - 125n + 720 = 0\\&\Rightarrow {n^2} - 25n + 144 = 0\\&\Rightarrow {n^2} - 16n - 9n + 144 = 0\\&\Rightarrow n\left( {n - 16} \right) - 9\left( {n - 16} \right) = 0\\&\Rightarrow \left( {n - 9} \right)\left( {n - 16} \right) = 0\\&\Rightarrow n = 9{\rm{ or}} 16\end{align}

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