NCERT Solutions For Class 11 Maths Chapter 9 Exercise 9.2

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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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