# Miscellaneous Exercise Sequences and Series - NCERT Class 11

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## Chapter 9 Ex.9.ME Question 1

Show that the sum of $${\left( {m + n} \right)^{th}}$$ and $${\left( {m - n} \right)^{th}}$$ terms of an A.P is equal to twice the $${m^{th}}$$term.

### Solution

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

It is known that the $${k^{th}}$$term of an A.P. is given by $${a_k} = a + \left( {k - 1} \right)d$$

Therefore,

\begin{align}{a_{m + n}}& = a + \left( {m + n - 1} \right)d\\{a_{m - n}} &= a + \left( {m - n - 1} \right)d\\{a_m}& = a + \left( {m - 1} \right)d\end{align}

Hence,

\begin{align}{a_{m + n}} + {a_{m - n}}& = a + \left( {m + n - 1} \right)d + a + \left( {m - n - 1} \right)d\\ &= 2a + \left( {m + n - 1 + m - n - 1} \right)d\\& = 2a + \left( {2m - 2} \right)d\\ &= 2a + 2\left( {m - 1} \right)d\\ &= 2\left[ {a + \left( {m - 1} \right)d} \right]\\& = 2{a_m}\end{align}

Thus, the sum of $${\left( {m + n} \right)^{th}}$$ and $${\left( {m - n} \right)^{th}}$$ terms of an A.P is equal to twice the $${m^{th}}$$ term.

## Chapter 9 Ex.9.ME Question 2

Let the sum of three numbers in A.P is $$24$$ and their product is $$440$$. Find the numbers.

### Solution

Let the three numbers in A.P be $$\left( {a - d} \right),a$$and $$\left( {a + d} \right)$$.

According to the given information,

\begin{align}&\left( {a - d} \right) + \left( a \right) + \left( {a - d} \right) = 24\\ &\Rightarrow\quad 3a = 24\\ &\Rightarrow\quad a = 8\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

And

\begin{align}&\left( {a - d} \right)\left( a \right)\left( {a - d} \right) = 440\\& \Rightarrow \left( {8 - d} \right)\left( 8 \right)\left( {8 + d} \right) = 440\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left[ {{\rm{from }}\left( 1 \right)} \right]\\&\Rightarrow \left( {8 - d} \right)\left( {8 + d} \right) = 55\\ &\Rightarrow 64 - {d^2} = 55\\ &\Rightarrow {d^2} = 9\\& \Rightarrow d = \pm 3\end{align}

Therefore,

when $$d = 3$$, the numbers are $$5,8$$and $$11$$

when $$d = - 3$$, the numbers are $$11, 8$$ and $$5$$.

Thus, the three numbers are $$5,8$$ and $$11$$.

## Chapter 9 Ex.9.ME Question 3

Let the sum of $$n,2n,3n$$ terms of an A.P be $${S_1},{S_2},{S_3}$$ respectively. Show that $${S_3} = 3\left( {{S_2} - {S_1}} \right)$$.

### Solution

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

Therefore,

\begin{align}{S_1} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\{S_2} &= \frac{{2n}}{2}\left[ {2a + \left( {2n - 1} \right)d} \right]\\ &= n\left[ {2a + \left( {2n - 1} \right)d} \right]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\{S_3} &= \frac{{3n}}{2}\left[ {2a + \left( {3n - 1} \right)d} \right]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}

By subtracting $$\left( 1 \right)$$ and , we obtain

\begin{align}{S_2} - {S_1}& = n\left[ {2a + \left( {2n - 1} \right)d} \right] - \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\ &= n\left[ {\frac{{4a + 4nd - 2d - 2a - nd + d}}{2}} \right]\\ &= n\left[ {\frac{{2a + 3nd - d}}{2}} \right]\\ &= \frac{n}{2}\left[ {2a + \left( {3n - 1} \right)d} \right]\\3\left( {{S_2} - {S_1}} \right) &= \frac{{3n}}{2}\left[ {2a + \left( {3n - 1} \right)d} \right]\\& = {S_3}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{from }}\left( 3 \right)} \right]\end{align}

Hence, $${S_3} = 3\left( {{S_2} - {S_1}} \right)$$ proved.

## Chapter 9 Ex.9.ME Question 4

Find the sum of all numbers between $$200$$ and $$400$$ which are divisible by $$7$$.

### Solution

The numbers lying between $$200$$ and $$400$$ which are divisible by $$7$$ are $$203,210,217, \ldots 399$$

Here $$a = 203,d = 7$$, and $${{a}_{n}}=399$$

Therefore,

\begin{align} {{a}_{n}}&=a+\left( n-1 \right)d \\ 399 &=203+\left( n-1 \right)7 \\ \left( n-1 \right)7&=196 \\ n-1&=28 \\ \Rightarrow n&=29 \\ \end{align}

Hence,

\begin{align}{S_{29}} &= \frac{{29}}{2}\left( {203 + 399} \right)\\ &= \frac{{29}}{2}\left( {602} \right)\\ &= 29 \times 301\\ &= 8729\end{align}

Thus, the required sum is $$8729$$.

## Chapter 9 Ex.9.ME Question 5

Find the sum of all integers from $$1$$ and $$100$$ that are divisible by $$2$$ or $$5$$.

### Solution

The integers from $$1$$ and $$100$$ that are divisible by $$2$$ are $$2,4,6, \ldots 100$$

This forms an A.P with both the first term and common difference equal to $$2$$.

Therefore,

\begin{align}{a_n} &= a + \left( {n - 1} \right)d\\100& = 2 + \left( {n - 1} \right)2\\100& = 2 + 2n - 2\\ \Rightarrow 2n &= 100\\ \Rightarrow n &= 50\end{align}

Therefore,

\begin{align}2 + 4 + 6 + \ldots + 100 &= \frac{{50}}{2}\left[ {2\left( 2 \right) + \left( {50 - 1} \right)\left( 2 \right)} \right]\\ &= \frac{{50}}{2}\left[ {4 + 98} \right]\\& = 25 \times 102\\ &= 2550\end{align}

The integers from $$1$$ to $$100$$ that are divisible by $$5$$ are $$5,10,15, \ldots 100$$

This forms an A.P with both the first term and common difference equal to $$5$$.

Therefore,

\begin{align}{a_n}& = a + \left( {n - 1} \right)d\\100 &= 5 + \left( {n - 1} \right)5\\100 &= 5 + 5n - 5\\ \Rightarrow 5n &= 100\\ \Rightarrow n &= 20\end{align}

Therefore,

\begin{align}5 + 10 + 15 + \ldots + 100 &= \frac{{20}}{2}\left[ {2\left( 5 \right) + \left( {20 - 1} \right)\left( 5 \right)} \right]\\ &= 10\left[ {10 + \left( {19} \right)5} \right]\\& = 10\left[ {10 + 95} \right]\\ A&= 10 \times 105\\ &= 1050\end{align}

The integers, which are divisible by both 2 and 5 are $$10,20,30, \ldots 100$$

This forms an A.P with both the first term and common difference equal to $$10$$.

Therefore,

\begin{align}100 &= 10 + \left( {n - 1} \right)10\\ \Rightarrow 10n &= 100\\ \Rightarrow n &= 10 \end{align}

Hence,

\begin{align}10 + 20 + \ldots + 100&= \frac{{10}}{2}\left[ {2\left( {10} \right) + \left( {10 - 1} \right)\left( {10} \right)} \right]\\ &= 5\left[ {20 + 90} \right]\\ &= 5 \times 110\\ &= 550\end{align}

Therefore, required sum $$= 2550 + 1050 - 550 = 3050$$

Thus, the sum of all integers from $$1$$ and $$100$$ that are divisible by $$2$$ or $$5$$ is $$3050$$.

## Chapter 9 Ex.9.ME Question 6

Find the sum of all two-digit numbers which when divided by $$4$$, yields $$1$$ as remainder.

### Solution

The two-digit numbers which when divided by $$4$$, yields $$1$$ as remainder are $$13,17,21, \ldots 97$$.

This forms an A.P with first term $$13$$ and common difference $$4$$.

Let $$n$$ be the number of terms of the A.P.

It is known that the $${n^{th}}$$ term of an A.P. is given by $${a^n} = a + \left( {n - 1} \right)d$$

Therefore,

\begin{align}97 &= 13 + \left( {n - 1} \right)\left( 4 \right)\\4\left( {n - 1} \right) &= 97 - 13\\n - 1 &= \frac{{84}}{4}\\& \Rightarrow n = 22\end{align}

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

Therefore,

\begin{align}{S_{22}}&= \frac{{22}}{2}\left[ {2\left( {13} \right) + \left( {22 - 1} \right)\left( 4 \right)} \right]\\&= 11\left[ {26 + 84} \right]\\&= 11 \times 110\\&= 1210\end{align}

Thus, the required sum is $$1210$$.

## Chapter 9 Ex.9.ME Question 7

If $$f$$ is a fraction satisfying $$f\left( {x + y} \right) = f\left( x \right) \cdot f\left( y \right)$$for all $$x,y \in N$$, such that $$f\left( 1 \right) = 3$$ and $$\sum\limits_{x = 1}^n {f\left( x \right)} = 120$$, find the value of $$n$$.

### Solution

$$f\left( {x + y} \right) = f\left( x \right) \cdot f\left( y \right)$$ for all $$x,y \in N$$ and $$f\left( 1 \right) = 3$$

Taking $$x = y = 1$$ in $$\left( 1 \right)$$, we obtain

\begin{align}f\left( {x + y} \right)&= f\left( x \right) \cdot f\left( y \right)\\f\left( {1 + 1} \right)&= f\left( 1 \right)f\left( 1 \right)\\f\left( 2 \right)&= 3 \times 3\\&= 9\end{align}

Similarly,

\begin{align}f\left( {x + y} \right)&= f\left( x \right) \cdot f\left( y \right)\\f\left( {1 + 2} \right)&= f\left( 1 \right)f\left( 2 \right)\\f\left( 3 \right) &= 3 \times 9\\ &= 27\end{align}

Also,

\begin{align}f\left( {x + y} \right)&= f\left( x \right) \cdot f\left( y \right)\\f\left( {1 + 3} \right)&= f\left( 1 \right)f\left( 3 \right)\\f\left( 4 \right)& = 3 \times 27\\& = 81\end{align}

Therefore, $$f\left( 1 \right),f\left( 2 \right),f\left( 3 \right)$$ i.e., $$3,9,27$$ forms a G.P. with both the first term and common ratio equal to $$3$$.

It is known that $${S_n} = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$$

It is given that $$\sum\limits_{k = 1}^n {f\left( x \right)} = 120$$

Therefore,

\begin{align}&\Rightarrow 120 = \frac{{3\left( {{3^n} - 1} \right)}}{{3 - 1}}\\&\Rightarrow 120 = \frac{3}{2}\left( {{3^n} - 1} \right)\\&\Rightarrow {3^n} - 1 = 80\\&\Rightarrow {3^n} = 81\\&\Rightarrow {3^n} = {3^4}\\&\Rightarrow n = 4\end{align}

Thus, the value of $$n = 4$$.

## Chapter 9 Ex.9.ME Question 8

The sum of some terms of G.P is $$315$$ whose first term and common ratio are $$5$$ and $$2$$ respectively. Find the last term and the number of terms.

### Solution

Let the sum of $$n$$ terms of the G.P be $$315$$.

It is known that $${S_n} = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$$

It is given that $$a = 5$$ and common ratio $$r = 2$$

Hence,

\begin{align}&315 = \frac{{5\left( {{2^n} - 1} \right)}}{{a^2 - 1}}\\&\Rightarrow {2^n} - 1 = 63\\&\Rightarrow {2^n} = 64 = {\left( 2 \right)^6}\\&\Rightarrow n = 6\end{align}

Therefore, last term of the G.P. is $$6^{th}$$ term

\begin{align}{a_6} &= a{r^{6 - 1}}\\& = a{r^5}\\ &= 5 \times {\left( 2 \right)^5}\\ &= 5 \times 32\\ &= 160\end{align}

Thus, the last term of the G.P is $$160$$.

## Chapter 9 Ex.9.ME Question 9

The first term of a G.P is $$1$$. The sum of the third term and fifth term is $$90$$. Find the common ratio of G.P.

### Solution

Let $$a$$ and $$r$$ be the first term and common ratio of the G.P respectively.

Hence,

\begin{align}a &= 1\\{a_3}& = a{r^2} = {r^2}\\{a_5}& = a{r^4} = {r^4}\end{align}

Therefore,

\begin{align}{r^2} + {r^4}& = 90\\{r^2} + {r^4} - 90 &= 0\\{r^2}&= \frac{{ - 1 + \sqrt {1 + 360} }}{2}\\ &= \frac{{ - 1 \pm \sqrt {361} }}{2}\\ &= \frac{{ - 1 \pm 19}}{2}\\ &= - 10\;or\;9\\{r^2} &= 9\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\text{Taking real roots}}} \right]\\r& = \pm 3\end{align}

Thus, the common ratio of the G.P is $$\pm 3$$.

## Chapter 9 Ex.9.ME Question 10

The sum of three numbers in G.P is $$56$$. If we subtract $$1,7,21$$ from these numbers in that order, we obtain an A.P. Find the numbers.

### Solution

Let the three numbers in G.P. be $$a,ar$$ and $$a{r^2}$$.

From the given condition,

\begin{align}&a + ar + a{r^2} = 56\\ &\Rightarrow\quad a\left( {1 + r + {r^2}} \right) = 56\;\;\;\;\;\;\;\;\ldots \left( 1 \right)\end{align}

Now, $$\left( {a - 1} \right),\left( {ar - 7} \right),\left( {a{r^2} - 21} \right)$$ forms an A.P.

Therefore,

\begin{align}&\left( {ar - 7} \right) - \left( {a - 1} \right)= \left( {a{r^2} - 21} \right) - \left( {ar - 7} \right)\\& \Rightarrow\quad ar - a - 6 = a{r^2} - ar - 14\\& \Rightarrow\quad a{r^2} - 2ar + a = 8\\ & \Rightarrow\quad a{r^2} - ar - ar + a = 8\\ & \Rightarrow\quad a\left( {{r^2} + 1 - 2r} \right) = 8\\ & \Rightarrow\quad 7a\left( {{r^2} - 2r + 1} \right) = 56\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

From and $$\left( 2 \right)$$, we get

\begin{align} & \Rightarrow 7\left( {{r}^{2}}-2r+1 \right)=\left( 1+r+{{r}^{2}} \right) \\ & \Rightarrow 7{{r}^{2}}-14r+7-1-r-{{r}^{2}}=0 \\ & \Rightarrow 6{{r}^{2}}-15r+6=0 \\ & \Rightarrow 2{{r}^{2}}-5r+2=0 \\ & \Rightarrow 2{{r}^{2}}-4r-r+2=0 \\ & \Rightarrow 2r\left( r-2 \right)-1\left( r-2 \right)=0 \\ & \Rightarrow \left( 2r-1 \right)\left( r-2 \right)=0 \\ & \Rightarrow r=\frac{1}{2},2 \\ \end{align}

$\left( 1 \right)$

Case I:

Substituting $$r = 2$$ in $$\left( 1 \right)$$, we obtain

\begin{align}&a\left( {1 + 2 + {2^2}} \right) = 56\\& \Rightarrow\quad 7a = 56\\& \Rightarrow\quad a = 8\end{align}

Hence, the three numbers are in G.P are $$8$$, $$16$$ and $$32$$.

Case II:

Substituting $$r = \frac{1}{2}$$ in $$\left( 1 \right)$$, we obtain

\begin{align}&a\left( {1 + \frac{1}{2} + {{\left( {\frac{1}{2}} \right)}^2}} \right) = 56\\ &\Rightarrow \quad \frac{7}{4}a = 56\\ &\Rightarrow\quad a = 32\end{align}

Hence, the three numbers are in G.P are $$32$$, $$16$$ and $$8$$.

Thus, the three required numbers are $$8$$, $$16$$ and $$32$$.

## Chapter 9 Ex.9.ME Question 11

A G.P consists of an even number of terms. If the sum of all terms is $$5$$ times the sum of terms occupying odd places, then find its common ratio.

### Solution

Let the G.P be $${T_1},{T_2},{T_3},{T_4}, \ldots {T_{2n}}$$.

According to the question,

\begin{align} &\Rightarrow {T_1} + {T_2} + {T_3} + {T_4} + \ldots + {T_{2n}} = 5\left[ {{T_1} + {T_3} + {T_5} + \ldots + {T_{2n - 1}}} \right]\\& \Rightarrow {T_1} + {T_2} + {T_3} + {T_4} + \ldots + {T_{2n}} - 5\left[ {{T_1} + {T_3} + {T_5} + \ldots + {T_{2n - 1}}} \right] = 0\\ &\Rightarrow {T_2} + {T_4} + \ldots + {T_{2n}} = 4\left[ {{T_1} + {T_3} + \ldots + {T_{2n - 1}}} \right]\end{align}

Let the G.P be $$a,ar,a{r^2}, \ldots$$

\begin{align}\frac{{ar\left( {{r^n} - 1} \right)}}{{r - 1}}& = \frac{{4 \times a\left( {{r^n} - 1} \right)}}{{r - 1}}\\ \Rightarrow\qquad ar &= 4a\\ \Rightarrow\qquad r &= 4\end{align}

Thus, the common ratio of the G.P is $$4$$.

## Chapter 9 Ex.9.ME Question 12

The sum of first four terms of an A.P is $$56$$. The sum of the last four terms is $$112$$. If its first term is $$11$$, then find the number of terms.

### Solution

Let the A.P be $$a,\left( {a + d} \right),\left( {a + 2d} \right),\left( {a + 3d} \right), \ldots a + \left( {n - 2} \right)d,a + \left( {n - 1} \right)d$$

It is given that, $$a = 11$$

Sum of the first four terms

\begin{align} &\Rightarrow a + \left( {a + d} \right) + \left( {a + 2d} \right) + \left( {a + 3d} \right) = 56\\& \Rightarrow 4a + 6d = 56\\ &\Rightarrow 4 \times 11 + 6d = 56\\ &\Rightarrow 6d = 56 - 44\\ &\Rightarrow d = \frac{{12}}{6}\\ &\Rightarrow d = 2\end{align}

Sum of last four terms

\begin{align} &\Rightarrow \left[ {a + \left( {n - 4} \right)d} \right] + \left[ {a + \left( {n - 3} \right)d} \right] + \left[ {a + \left( {n - 2} \right)d} \right] + \left[ {a + \left( {n - 1} \right)d} \right] = 112\\ &\Rightarrow 4a + \left( {4n - 10} \right)d = 112\\ &\Rightarrow 4 \times 11 + \left( {4n - 10} \right) \times 2 = 112\\ &\Rightarrow 44 + 8n - 20 = 112\\ &\Rightarrow 8n = 112 - 24\\ &\Rightarrow n = \frac{{88}}{8}\\ &\Rightarrow n = 11\end{align}

Thus, the number of terms of the A.P is $$11$$.

## Chapter 9 Ex.9.ME Question 13

If $$\frac{{a + bx}}{{a - bx}} = \frac{{b + cx}}{{b - cx}} = \frac{{c + dx}}{{c - dx}}\left( {x \ne 0} \right),$$ then show that $$a,b,c$$ and $$d$$ are in G.P.

### Solution

It is given that $$\frac{{a + bx}}{{a - bx}} = \frac{{b + cx}}{{b - cx}} = \frac{{c + dx}}{{c - dx}}\left( {x \ne 0} \right),$$

Therefore,

\begin{align} &\Rightarrow \frac{{a + bx}}{{a - bx}} = \frac{{b + cx}}{{b - cx}}\\& \Rightarrow \left( {a + bx} \right)\left( {b - cx} \right) = \left( {b + cx} \right)\left( {a - bx} \right)\\& \Rightarrow ab - acx + {b^2}x - bc{x^2} = ab - {b^2}x + acx - bc{x^2}\\ &\Rightarrow 2{b^2}x = 2acx\\ &\Rightarrow {b^2} = ac\\ &\Rightarrow \frac{b}{a} = \frac{c}{b}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Also,

\begin{align}& \Rightarrow \frac{{b + cx}}{{b - cx}} = \frac{{c + dx}}{{c - dx}}\\ &\Rightarrow \left( {b + cx} \right)\left( {c - dx} \right) = \left( {b - cx} \right)\left( {c + dx} \right)\\ &\Rightarrow bc - bdx + {c^2}x - cd{x^2} = bc + bdx - {c^2}x - cd{x^2}\\ &\Rightarrow 2{c^2}x = 2bdx\\ &\Rightarrow {c^2} = bd\\ &\Rightarrow \frac{c}{b} = \frac{d}{c}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

From $$\left( 1 \right)$$ and $$\left( 2 \right)$$, we obtain

$\frac{b}{a} = \frac{c}{b} = \frac{d}{c}$

Thus, $$a,b,c$$ and $$d$$ are in G.P.

## Chapter 9 Ex.9.ME Question 14

Let $$S$$ be the sum, $$P$$ the product and $$R$$ the sum of reciprocals of $$n$$ terms in a G.P. Prove that $${P^2}{R^n} = {S^n}$$.

### Solution

Let the G.P be $$a,ar,a{r^2},a{r^3} \ldots a{r^{n - 1}}$$

According to the given information

Sum of the $$n$$ terms

$S = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$

Product of the $$n$$ terms

\begin{align} P &={{a}^{n}}\times {{r}^{1+2+3+\ldots +n-1}} \\ & ={{a}^{n}}{{r}^{\frac{n\left( n-1 \right)}{2}}}\ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \text{sum of first }n\text{ natural numbers }=n\frac{\left( n+1 \right)}{2} \right] \end{align}

Sum of reciprocals of $$n$$ terms

\begin{align}& R=\frac{1}{a}+\frac{1}{ar}+\ldots +\frac{1}{a{{r}^{n-1}}} \\ & =\frac{{{r}^{n-1}}+{{r}^{n-2}}+\ldots r+1}{a{{r}^{n-1}}} \\ & =\frac{1\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}\times \frac{1}{a{{r}^{n-1}}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because 1,r,\ldots {{r}^{n-1}}\text{ forms a G}\text{.P} \right] \\ & =\frac{{{r}^{n}}-1}{a{{r}^{n-1}}\left( r-1 \right)}\end{align}

Therefore,

\begin{align}{P^2}{R^n} &= {a^{2n}}{r^{n\left( {n - 1} \right)}}\frac{{{{\left( {{r^n} - 1} \right)}^n}}}{{{a^n}{r^{n\left( {n - 1} \right)}}{{\left( {r - 1} \right)}^n}}}\\ &= \frac{{{a^n}{{\left( {{r^n} - 1} \right)}^n}}}{{{{\left( {r - 1} \right)}^n}}}\\ &= {\left[ {\frac{{a\left( {{r^n} - 1} \right)}}{{\left( {r - 1} \right)}}} \right]^n}\\ &= {S^n}\end{align}

Hence, $${P^2}{R^n} = {S^n}$$ proved

## Chapter 9 Ex.9.ME Question 15

The $${p^{th}},{q^{th}}$$ and $${r^{th}}$$ terms of an A.P are $$a,b,c$$ respectively. Show that

$$\left( {q - r} \right)a + \left( {r - p} \right)b + \left( {p - q} \right)c = 0$$.

### Solution

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

The $${n^{th}}$$ term of an A.P is given by, $${a_n} = t + \left( {n - 1} \right)d$$

\begin{align}{a_p} &= t + \left( {p - 1} \right)d = a\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\ {a_q}& = t + \left( {q - 1} \right)d = b\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\{a_r} &= t + \left( {r - 1} \right)d = c\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}

Subtracting ($$2$$) from ($$1$$), we obtain

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

Subtracting ($$3$$) from ($$2$$), we obtain

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

From ($$4$$) and ($$5$$), we obtain

\begin{align} &\Rightarrow \frac{{a - b}}{{p - q}} = \frac{{b - c}}{{q - r}}\\ &\Rightarrow \left( {a - b} \right)\left( {q - r} \right) = \left( {b - c} \right)\left( {p - q} \right)\\ &\Rightarrow aq - ar - bq + br = bp - bq - cp + cq\\ &\Rightarrow bp - cp + cq - aq + ar - br = 0\\& \Rightarrow \left( { - aq + ar} \right) + \left( {bp - br} \right) + \left( { - cp + cq} \right) = 0\\& \Rightarrow - a\left( {q - r} \right) - b\left( {r - p} \right) - c\left( {p - q} \right) = 0\\& \Rightarrow a\left( {q - r} \right) + b\left( {r - p} \right) + c\left( {p - q} \right) = 0\end{align}

Hence, $$\left( {q - r} \right)a + \left( {r - p} \right)b + \left( {p - q} \right)c = 0$$ proved.

## Chapter 9 Ex.9.ME Question 16

If $$a\left( \frac{1}{b}+\frac{1}{c} \right),b\left( \frac{1}{c}+\frac{1}{a} \right),c\left( \frac{1}{a}+\frac{1}{b} \right)$$ are in A.P, prove that $$a,b,c$$ are in A.P.

### Solution

It is given that $$a\left( {\frac{1}{b} + \frac{1}{c}} \right),b\left( {\frac{1}{c} + \frac{1}{a}} \right),c\left( {\frac{1}{a} + \frac{1}{b}} \right)$$are in A.P.

Therefore,

\begin{align} &\Rightarrow b\left( {\frac{1}{c} + \frac{1}{a}} \right) - a\left( {\frac{1}{b} + \frac{1}{c}} \right) = c\left( {\frac{1}{a} + \frac{1}{b}} \right) - b\left( {\frac{1}{c} + \frac{1}{a}} \right)\\ &\Rightarrow \frac{{b\left( {a + c} \right)}}{{ac}} - \frac{{a\left( {b + c} \right)}}{{bc}} = \frac{{c\left( {a + b} \right)}}{{ab}} - \frac{{b\left( {a + c} \right)}}{{ac}}\\& \Rightarrow \frac{{{b^2}a + {b^2}c - {a^2}b - {a^2}c}}{{abc}} = \frac{{{c^2}a + {c^2}b - {b^2}a - {b^2}c}}{{abc}}\\ &\Rightarrow {b^2}a - {a^2}b + {b^2}c - {a^2}c = {c^2}a - {b^2}a + {c^2}b - {b^2}c\\ &\Rightarrow ab\left( {b - a} \right) + c\left( {{b^2} - {a^2}} \right) = a\left( {{c^2} - {b^2}} \right) + bc\left( {c - b} \right)\\& \Rightarrow ab\left( {b - a} \right) + c\left( {b + a} \right)\left( {b - a} \right) = a\left( {c - b} \right)\left( {c + b} \right) + bc\left( {c - b} \right)\\& \Rightarrow \left( {b - a} \right)\left( {ab + cb + ca} \right) = \left( {c - b} \right)\left( {ac + ab + bc} \right)\\& \Rightarrow b - a = c - b\end{align}

Thus, $$a,b,c$$ are in A.P. proved.

## Chapter 9 Ex.9.ME Question 17

If $$a,b,c,d$$ are in G.P, prove that $$\left( {{a^n} + {b^n}} \right),\left( {{b^n} + {c^n}} \right),\left( {{c^n} + {d^n}} \right)$$ are in G.P.

### Solution

It is given that $$a,b,c,d$$ are in G.P.

Therefore,

\begin{align}{b^2}& = ac\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\{c^2} &= bd\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\ad &= bc\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}

We need to prove $$\left( {{a^n} + {b^n}} \right),\left( {{b^n} + {c^n}} \right),\left( {{c^n} + {d^n}} \right)$$ are in G.P. i.e.,

${\left( {{b^n} + {c^n}} \right)^2} = \left( {{a^n} + {b^n}} \right) \cdot \left( {{c^n} + {d^n}} \right)$

Consider

\begin{align}{\left( {{b^n} + {c^n}} \right)^2}& = {b^{2n}} + 2{b^n}{c^n} + {c^{2n}}\\& = {\left( {{b^2}} \right)^n} + 2{b^n}{c^n} + {\left( {{c^2}} \right)^n}\\ &= {\left( {ac} \right)^n} + 2{b^n}{c^n} + {\left( {bd} \right)^n}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{Using }}\left( 1 \right){\rm{ and }}\left( 2 \right)} \right]\\ &= {a^n}{c^n} + {b^n}{c^n} + {b^n}{c^n} + {b^n}{d^n}\\ &= {a^n}{c^n} + {b^n}{c^n} + {a^n}{d^n} + {b^n}{d^n}\;\;\;\;\;\;\;\;\;\left[ {{\rm{Using }}\left( 3 \right)} \right]\\ &= {c^n}\left( {{a^n} + {b^n}} \right) + {d^n}\left( {{a^n} + {b^n}} \right)\\ &= \left( {{a^n} + {b^n}} \right)\left( {{c^n} + {d^n}} \right)\end{align}

Hence, $${\left( {{b^n} + {c^n}} \right)^2} = \left( {{a^n} + {b^n}} \right) \cdot \left( {{c^n} + {d^n}} \right)$$

Thus, $$\left( {{a^n} + {b^n}} \right),\left( {{b^n} + {c^n}} \right),\left( {{c^n} + {d^n}} \right)$$ are in G.P.

## Chapter 9 Ex.9.ME Question 18

If $$a$$ and $$b$$ are the roots of $${x^2} - 3x + p = 0$$ and $$c,\;d$$ are the roots of $${x^2} - 12x + q = 0$$ where $$a,b,c,d$$ form a G.P. Prove that $$\left( {q + p} \right):\left( {q - p} \right) = 17:15.$$

### Solution

It is given that $$a$$ and $$b$$are the roots of $${x^2} - 3x + p = 0$$

Therefore,

$a + b = 3,\;\;\;\;\;\;ab = p\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)$

Also, $$c$$ and $$d$$ are the roots of $${x^2} - 12x + q = 0$$

$c + d = 12,\;\;\;\;\;\;cd = q\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)$

It is given that $$a,b,c,d$$ form a G.P.

Let $$a = x,b = xr,c = x{r^2},d = x{r^3}$$

Form $$\left( 1 \right)$$ and $$\left( 2 \right)$$, we obtain

\begin{align} &\Rightarrow x + xr = 3\\& \Rightarrow x\left( {1 + r} \right) = 3\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\\\\& \Rightarrow x{r^2} + x{r^3} = 12\\& \Rightarrow x{r^2}\left( {1 + r} \right) = 12\;\;\;\;\;\;\;\;\; \ldots \left( 4 \right)\end{align}

On dividing $$\left( 4 \right)$$ by $$\left( 3 \right)$$, we obtain

\begin{align}\frac{{x{r^2}\left( {1 + r} \right)}}{{x\left( {1 + r} \right)}}& = \frac{{12}}{3}\\ \Rightarrow {r^2} &= 4\\ \Rightarrow r &= \pm 2\end{align}

Case I: when $$r = 2,\;x = 1$$

\begin{align}ab &= {x^2}r = 2,\\cd &= {x^2}{r^5} = 32\end{align}

Therefore,

\begin{align}& \Rightarrow \frac{{q + p}}{{q - p}} = \frac{{32 + 2}}{{32 - 2}} = \frac{{34}}{{30}} = \frac{{17}}{{15}}\\& \Rightarrow \left( {q + p} \right):\left( {q - p} \right) = 17:15\end{align}

Case II: when $$r = - 2,\;x = - 3$$

\begin{align}ab &= {x^2}r = - 18\\cd &= {x^2}{r^5} = - 288\end{align}

Therefore,

\begin{align} &\Rightarrow \frac{{q + p}}{{q - p}} = \frac{{ - 288 - 18}}{{ - 288 + 18}} = \frac{{ - 306}}{{ - 270}} = \frac{{17}}{{15}}\\ &\Rightarrow \left( {q + p} \right):\left( {q - p} \right) = 17:15\end{align}

Thus, $$\left( {q + p} \right):\left( {q - p} \right) = 17:15$$ proved

## Chapter 9 Ex.9.ME Question 19

The ratio of the A.M and G.M of two positive numbers $$a$$ and $$b$$, is $$m:n$$. Show that $$a:b = \left( {m + \sqrt {{m^2} - {n^2}} } \right):\left( {m - \sqrt {{m^2} - {n^2}} } \right)$$.

### Solution

Let the two numbers be $$a$$ and $$b$$.

\begin{align}A.M &= \frac{{a + b}}{2}\\\\G.M &= \sqrt {ab} \end{align}

According to the given condition,

\begin{align} &\Rightarrow \frac{{a + b}}{{2\sqrt {ab} }} = \frac{m}{n}\\ &\Rightarrow \frac{{{{\left( {a + b} \right)}^2}}}{{4ab}} = \frac{{{m^2}}}{{{n^2}}}\\& \Rightarrow {\left( {a + b} \right)^2} = \frac{{4ab{m^2}}}{{{n^2}}}\\ &\Rightarrow \left( {a + b} \right) = \frac{{2m\sqrt {ab} }}{n}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Using this in the identity $${\left( {a - b} \right)^2} = {\left( {a + b} \right)^2} - 4ab$$, we obtain

\begin{align}{\left( {a - b} \right)^2} &= \frac{{4ab{m^2}}}{{{n^2}}} - 4ab\\ &= \frac{{4ab\left( {{m^2} - {n^2}} \right)}}{{{n^2}}}\\\left( {a - b} \right) &= \frac{{2\sqrt {ab} \sqrt {{m^2} - {n^2}} }}{n}\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Adding ($$1$$) and ($$2$$), we obtain

\begin{align}2a &= \frac{{2\sqrt {ab} }}{n}\left( {m + \sqrt {{m^2} - {n^2}} } \right)\\a &= \frac{{\sqrt {ab} }}{n}\left( {m + \sqrt {{m^2} - {n^2}} } \right)\end{align}

Substituting the value of $$a$$ in $$\left( {\rm{1}} \right)$$, we obtain

\begin{align}b &= \frac{{2\sqrt {ab} }}{n}m - \frac{{\sqrt {ab} }}{n}\left( {m + \sqrt {{m^2} - {n^2}} } \right)\\ &= \frac{{\sqrt {ab} }}{n}m - \frac{{\sqrt {ab} }}{n}\left( {\sqrt {{m^2} - {n^2}} } \right)\\ &= \frac{{\sqrt {ab} }}{n}\left( {m - \sqrt {{m^2} - {n^2}} } \right)\end{align}

Therefore,

\begin{align}\frac{a}{b} &= \frac{{\left( {\frac{{\sqrt {ab} }}{n}\left( {m + \sqrt {{m^2} - {n^2}} } \right)} \right)}}{{\left( {\frac{{\sqrt {ab} }}{n}\left( {m - \sqrt {{m^2} - {n^2}} } \right)} \right)}}\\ &= \frac{{\left( {m + \sqrt {{m^2} - {n^2}} } \right)}}{{\left( {m - \sqrt {{m^2} - {n^2}} } \right)}}\end{align}

Thus, $$a:b = \left( {m + \sqrt {{m^2} - {n^2}} } \right):\left( {m - \sqrt {{m^2} - {n^2}} } \right)$$.

## Chapter 9 Ex.9.ME Question 20

If $$a,b,c$$ are in A.P;$$b,c,d$$ are in G.P and $$\frac{1}{c},\frac{1}{d},\frac{1}{e}$$ are in A.P. Prove that  are in G.P.

### Solution

It is given that $$a,b,c$$ are in A.P

\begin{align}b - a &= c - b\\b &= \frac{{a + c}}{2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

It is given that $$b,c,d$$ are in G.P

\begin{align}{c^2} &= bd\\{c^2} &= \left( {\frac{{a + c}}{2}} \right)d\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{from }}\left( 1 \right)} \right]\\d &= \frac{{2{c^2}}}{{a + c}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Also, $$\frac{1}{c},\frac{1}{d},\frac{1}{e}$$are in A.P.

Therefore,

\begin{align}& \Rightarrow \frac{1}{d} - \frac{1}{c} = \frac{1}{e} - \frac{1}{d}\\ & \Rightarrow \frac{2}{d} = \frac{1}{c} + \frac{1}{e}\\ &\Rightarrow \frac{2}{{\frac{{2{c^2}}}{{a + c}}}} = \frac{1}{c} + \frac{1}{e}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{from }}\left( 2 \right)} \right]\\& \Rightarrow \frac{{2\left( {a + c} \right)}}{{2{c^2}}} = \frac{1}{c} + \frac{1}{e}\\ &\Rightarrow \frac{{\left( {a + c} \right)}}{{{c^2}}} =\frac{{e + c}}{{ce}}\\ &\Rightarrow \frac{{\left( {a + c} \right)}}{c} = \frac{{e + c}}{e}\\ &\Rightarrow \left( {a + c} \right)e = \left( {e + c} \right)c\\ &\Rightarrow ae + ce = ec + {c^2}\\ &\Rightarrow {c^2} = ae\end{align}

Thus, $$a,c,e$$ are in G.P.

## Chapter 9 Ex.9.ME Question 21

Find the sum of the following series up to $$n$$ terms:

(i) $$5 + 55 + 555 + \ldots$$

(ii) $$.6 + .66 + .666 + \ldots$$

### Solution

(i) $$5 + 55 + 555 + \ldots$$

Let$${S_n} = 5 + 55 + 555 + \ldots n{\rm{ terms}}$$

\begin{align}{S_n}& = \frac{5}{9}\left[ {9 + 99 + 999 + \ldots {\rm{ }}n{\rm{ terms}}} \right]\\ &= \frac{5}{9}\left[ {\left( {10 - 1} \right) + \left( {{{10}^2} - 1} \right) + \left( {{{10}^3} - 1} \right) + \ldots {\rm{ }}n{\rm{ terms}}} \right]\\ &= \frac{5}{9}\left[ {\left( {10 + {{10}^2} + {{10}^3} + \ldots {\rm{ }}n{\rm{ terms}}} \right) - \left( {1 + 1 + \ldots {\rm{ }}n{\rm{ terms}}} \right)} \right]\\ &= \frac{5}{9}\left[ {\frac{{10\left( {{{10}^n} - 1} \right)}}{{10 - 1}} - n} \right]\\ &= \frac{5}{9}\left[ {\frac{{10\left( {{{10}^n} - 1} \right)}}{9} - n} \right]\\ &= \frac{{50}}{{81}}\left( {{{10}^n} - 1} \right) - \frac{{5n}}{9}\end{align}

(ii) $$.6 + .66 + .666 + \ldots$$

Let$${S_n} = .6 + .66 + .666 + \ldots {\rm{ }}n{\rm{ terms}}$$

\begin{align}{S_n}& = 6\left[ {0.1 + 0.11 + 0.111 + \ldots {\rm{ }}n{\rm{ terms}}} \right]\\ &= \frac{6}{9}\left[ {0.9 + 0.99 + 0.999 + \ldots {\rm{ }}n{\rm{ terms}}} \right]\\ &= \frac{6}{9}\left[ {\left( {1 - \frac{1}{{10}}} \right) + \left( {1 - \frac{1}{{{{10}^2}}}} \right) + \left( {1 - \frac{1}{{{{10}^3}}}} \right) + \ldots {\rm{ }}n{\rm{ terms}}} \right]\\ &= \frac{2}{3}\left[ {\left( {1 + 1 + 1 + \ldots {\rm{ }}n{\rm{ terms}}} \right) - \frac{1}{{10}}\left( {1 + \frac{1}{{10}} + \frac{1}{{{{10}^2}}} + \ldots n{\rm{ terms}}} \right)} \right]\\ &= \frac{2}{3}\left[ {n - \frac{1}{{10}}\left( {\frac{{1 - {{\left( {\frac{1}{{10}}} \right)}^n}}}{{1 - \frac{1}{{10}}}}} \right)} \right]\\ &= \frac{2}{3}n - \frac{2}{{30}} \times \frac{{10}}{9}\left( {1 - {{10}^{ - n}}} \right)\\& = \frac{2}{3}n - \frac{2}{{27}}\left( {1 - {{10}^{ - n}}} \right)\end{align}

## Chapter 9 Ex.9.ME Question 22

Find the $${20^{th}}$$ term of the series $$2 \times 4 + 4 \times 6 + 6 \times 8 + \ldots + n{\rm{ terms}}$$.

### Solution

The given series is$$2 \times 4 + 4 \times 6 + 6 \times 8 + \ldots + n{\rm{ terms}}$$

Therefore,

\begin{align}{a_n} = 2n \times \left( {2n + 2} \right)\\ = 4{n^2} + 4n\end{align}

\begin{align}{a_{20}}& = 4{\left( {20} \right)^2} + 4\left( {20} \right)\\ &= 4\left( {400} \right) + 80\\& = 1680\end{align}

Thus, the $${20^{th}}$$ term of the series is $$1680$$.

## Chapter 9 Ex.9.ME Question 23

Find the sum of the first $$n$$ terms of the series: $$3 + 7 + 13 + 21 + 31 + \ldots$$

### Solution

The given series is $$3 + 7 + 13 + 21 + 31 + \ldots$$, can be written as

\begin{align}S &= 3 + 7 + 13 + 21 + 31 + \ldots + {a_{n - 1}} + {a_n}\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\S &= 3 + 7 + 13 + 21 + \ldots + {a_{n - 2}} + {a_{n - 1}} + {a_n}\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

On subtracting the equation $$\left( 2 \right)$$ from $$\left( 1 \right)$$, we obtain

\begin{align}S - S &= \left[ {3 + 7 + 13 + 21 + 31 + \ldots + {a_{n - 1}} + {a_n}} \right] - \left[ {3 + 7 + 13 + 21 + \ldots + {a_{n - 2}} + {a_{n - 1}} + {a_n}} \right]\\0 &= 3 + \left( {7 + 13 + 21 + 31 + \ldots + {a_{n - 1}} + {a_n}} \right) - \left( {3 + 7 + 13 + 21 + \ldots + {a_{n - 2}} + {a_{n - 1}}} \right) - {a_n}\\0 &= 3 + \left[ {\left( {7 - 3} \right) + \left( {13 - 7} \right) + \left( {21 - 13} \right) + \ldots + \left( {{a_n} - {a_{n - 1}}} \right)} \right] - {a_n}\\0 &= 3 + \left[ {4 + 6 + 8 + \ldots \left( {n - 1} \right){\rm{ terms}}} \right] - {a_n}\\{a_n} &= 3 + \left[ {4 + 6 + 8 + \ldots \left( {n - 1} \right){\rm{ terms}}} \right]\\ &= 3 + \left( {\frac{{n - 1}}{2}} \right)\left[ {2 \times 4 + \left( {n - 1 - 1} \right)2} \right]\\& = 3 + \left( {\frac{{n - 1}}{2}} \right)\left[ {8 + 2n - 4} \right]\\ &= 3 + \left( {\frac{{n - 1}}{2}} \right)\left[ {2n + 4} \right]\\ &= 3 + \left( {n - 1} \right)\left( {n + 2} \right)\\ &= 3 + \left( {{n^2} + n - 2} \right)\\{a_n} &= {n^2} + n + 1\end{align}

Therefore,

\begin{align}\sum\limits_{k = 1}^n {{a_k}} &= \sum\limits_{k = 1}^n {{k^2} + k + 1} \\ &= \sum\limits_{k = 1}^n {{k^2} + \sum\limits_{k = 1}^n {k + } \sum\limits_{k = 1}^n 1 } \\ &= \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \frac{{n\left( {n + 1} \right)}}{2} + n\\ &= n\left[ {\frac{{\left( {n + 1} \right)\left( {2n + 1} \right) + 3\left( {n + 1} \right) + 6}}{6}} \right]\end{align}

\begin{align}\sum\limits_{k = 1}^n {{a_k}} &= n\left[ {\frac{{2{n^2} + 3n + 1 + 3n + 3 + 6}}{6}} \right]\\ &= n\left[ {\frac{{2{n^2} + 6n + 10}}{6}} \right]\\ &= \frac{n}{3}\left( {{n^2} + 3n + 5} \right)\end{align}

## Chapter 9 Ex.9.ME Question 24

If $${S_1},{S_2},{S_3}$$ are the sums of first $$n$$ natural numbers, their squares and their cubes respectively, show that $$9S_2^2 = {S_3}\left( {1 + 8{S_1}} \right)$$.

### Solution

From the given information,

\begin{align}{S_1} &= \frac{{n\left( {n + 1} \right)}}{2}\\{S_2} &= \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\\{S_3} &= {\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2}\end{align}

Hence,

\begin{align}9{S_2}^2 &= 9{\left[ {\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right]^2}\\ &= \frac{9}{{36}}{\left[ {n\left( {n + 1} \right)\left( {2n + 1} \right)} \right]^2}\\ &= \frac{1}{4}{\left[ {n\left( {n + 1} \right)\left( {2n + 1} \right)} \right]^2}\\ &= {\left[ {\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{2}} \right]^2}\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Also,

\begin{align}{S_3}\left( {1 + 8{S_1}} \right) &= {\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2}\left[ {1 + \frac{{8n\left( {n + 1} \right)}}{2}} \right]\\ &= {\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2}\left[ {1 + 4{n^2} + 4n} \right]\\ &= {\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2}{\left( {2n + 1} \right)^2}\\ &= {\left[ {\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{2}} \right]^2}\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

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

$9S_2^2 = {S_3}\left( {1 + 8{S_1}} \right)$

## Chapter 9 Ex.9.ME Question 25

Find the sum of the following series up to $$n$$ terms:

$\frac{{{1^3}}}{1} + \frac{{{1^3} + {2^3}}}{{1 + 3}} + \frac{{{1^3} + {2^3} + {3^3}}}{{1 + 3 + 5}} + \ldots$

### Solution

The $${n^{th}}$$term of the given series $$\frac{{{1^3}}}{1} + \frac{{{1^3} + {2^3}}}{{1 + 3}} + \frac{{{1^3} + {2^3} + {3^3}}}{{1 + 3 + 5}} + \ldots$$ is

\begin{align}{a_n} &= \frac{{{1^3} + {2^3} + {3^3} + \ldots {n^3}}}{{1 + 3 + 5 + \ldots + \left( {2n - 1} \right)}}\\ &= \frac{{{{\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]}^2}}}{{1 + 3 + 5 + \ldots + \left( {2n - 1} \right)}}\end{align}

Here, $${\rm{1,3,5,}} \ldots \left( {2n - 1} \right)$$ is an A.P. with first term $$a$$, last term $$\left( {2n - 1} \right)$$ and number of terms $$n$$.

Therefore,

\begin{align}1 + 3 + 5 + \dots + \left( {2n - 1} \right) &= \frac{n}{2}\left[ {2 \times 1 + \left( {n - 1} \right)2} \right]\\ &= {n^2}\end{align}

Hence,

\begin{align}{a_n} &= \frac{{{n^2}{{\left( {n + 1} \right)}^2}}}{{4{n^2}}}\\ &= \frac{{{n^2} + 2n + 1}}{4}\\ &= \frac{1}{4}{n^2} + \frac{1}{2}n + \frac{1}{4}\end{align}

Thus,

\begin{align}{S_n} &= \sum\limits_{k = 1}^n {{a_k}} \\ Z&= \sum\limits_{k = 1}^n {\left( {\frac{1}{4}{k^2} + \frac{1}{2}k + \frac{1}{4}} \right)} \\ &= \frac{1}{4}\left[ {\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right] + \frac{1}{2}\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right] + \frac{1}{4}n\\ &= \frac{{n\left[ {\left( {n + 1} \right)\left( {2n + 1} \right) + 6\left( {n + 1} \right) + 6} \right]}}{{24}}\\& = \frac{{n\left[ {2{n^2} + 3n + 1 + 6n + 6 + 6} \right]}}{{24}}\\ &= \frac{n}{{24}}\left( {2{n^2} + 9n + 13} \right)\end{align}

## Chapter 9 Ex.9.ME Question 26

Show that $$\frac{{1 \times {2^2} + 2 \times {3^2} + \ldots + n \times {{\left( {n + 1} \right)}^2}}}{{{1^2} \times 2 + {2^2} \times 3 + \ldots + {n^2} \times \left( {n + 1} \right)}} = \frac{{3n + 5}}{{3n + 1}}$$.

### Solution

The $${n^{th}}$$ term of the given series $$1 \times {2^2} + 2 \times {3^2} + \ldots + n \times {\left( {n + 1} \right)^2}$$ in numerator

\begin{align}{a_n} &= n{\left( {n + 1} \right)^2}\\ &= {n^3} + 2{n^2} + n\end{align}

The $${n^{th}}$$ term of the given series $${1^2} \times 2 + {2^2} \times 3 + \ldots + {n^2} \times \left( {n + 1} \right)$$ in denominator

\begin{align}{a_n} &= {n^2}\left( {n + 1} \right)\\& = {n^3} + {n^2}\end{align}

Therefore,

\begin{align}\frac{{1 \times {2^2} + 2 \times {3^2} + \ldots + n \times {{\left( {n + 1} \right)}^2}}}{{{1^2} \times 2 + {2^2} \times 3 + \ldots + {n^2} \times \left( {n + 1} \right)}} &= \frac{{\sum\limits_{k = 1}^n {{a_k}} }}{{\sum\limits_{k = 1}^n {{a_k}} }}\\ &= \frac{{\sum\limits_{k = 1}^n {\left( {{k^3} + 2{k^2} + k} \right)} }}{{\sum\limits_{k = 1}^n {\left( {{k^3} + {k^2}} \right)} }}\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Here,

\begin{align}\sum\limits_{k = 1}^n {\left( {{k^3} + 2{k^2} + k} \right)} &= \frac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} + \frac{{2n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \frac{{n\left( {n + 1} \right)}}{2}\\ &= \frac{{n\left( {n + 1} \right)}}{2}\left[ {\frac{{n\left( {n + 1} \right)}}{2} + \frac{2}{3}\left( {2n + 1} \right) + 1} \right]\\ &= \frac{{n\left( {n + 1} \right)}}{2}\left[ {\frac{{3{n^2} + 3n + 8n + 4 + 6}}{6}} \right]\\ &= \frac{{n\left( {n + 1} \right)}}{{12}}\left[ {3{n^2} + 11n + 10} \right]\\& = \frac{{n\left( {n + 1} \right)}}{{12}}\left[ {3{n^2} + 6n + 5n + 10} \right]\\& = \frac{{n\left( {n + 1} \right)}}{{12}}\left[ {3{n^2}\left( {n + 2} \right) + 5\left( {n + 2} \right)} \right]\\ &= \frac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {3n + 5} \right)}}{{12}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Also,

\begin{align}\sum\limits_{k = 1}^n {\left( {{k^3} + {k^2}} \right)} &= \frac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} + \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\\& = \frac{{n\left( {n + 1} \right)}}{2}\left[ {\frac{{n\left( {n + 1} \right)}}{2} + \frac{{2n + 1}}{3}} \right]\\ &= \frac{{n\left( {n + 1} \right)}}{2}\left[ {\frac{{3{n^2} + 3n + 4n + 2}}{6}} \right]\\ &= \frac{{n\left( {n + 1} \right)}}{{12}}\left[ {3{n^2} + 7n + 2} \right]\\ &= \frac{{n\left( {n + 1} \right)}}{{12}}\left[ {3{n^2} + 6n + n + 2} \right]\\ &= \frac{{n\left( {n + 1} \right)}}{{12}}\left[ {3n\left( {n + 2} \right) + 1\left( {n + 2} \right)} \right]\\ &= \frac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {3n + 1} \right)}}{{12}}\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}

From $$\left( {\rm{1}} \right){\rm{,}}\left( {\rm{2}} \right)$$and $$\left( 3 \right)$$, we obtain

\begin{align}\frac{{1 \times {2^2} + 2 \times {3^2} + \ldots + n \times {{\left( {n + 1} \right)}^2}}}{{{1^2} \times 2 + {2^2} \times 3 + \ldots + {n^2} \times \left( {n + 1} \right)}} &= \frac{{\frac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {3n + 5} \right)}}{{12}}}}{{\frac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {3n + 1} \right)}}{{12}}}}\\ &= \frac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {3n + 5} \right)}}{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {3n + 1} \right)}}\\ &= \frac{{3n + 5}}{{3n + 1}}\end{align}

Thus, $$\frac{{1 \times {2^2} + 2 \times {3^2} + \ldots + n \times {{\left( {n + 1} \right)}^2}}}{{{1^2} \times 2 + {2^2} \times 3 + \ldots + {n^2} \times \left( {n + 1} \right)}} = \frac{{3n + 5}}{{3n + 1}}$$ proved.

## Chapter 9 Ex.9.ME Question 27

A farmer buys a used tractor for $$₹$$ $$12000$$. He pays $$₹$$ $$6000$$ cash and agrees to pay the balance in annual instalment of $$₹$$ $$500$$ plus $$12\%$$interest on the unpaid amount. How much will the tractor cost him?

### Solution

It is given that farmer pays $$₹$$ $$6000$$ in cash.

Hence, unpaid amount in $$₹$$ $$= {\rm{12}}000 - 6000 = 6000$$

According to the given condition, the interest paid annually is

$\left( {12\% {\rm{ }}of{\rm{ }}6000} \right),\left( {12\% {\rm{ }}of{\rm{ }}5500} \right),\left( {12\% {\rm{ }}of{\rm{ }}5000} \right), \ldots \left( {12\% {\rm{ }}of{\rm{ }}500} \right)$

Thus, total interest to be paid$$500 + 1000 + \ldots + 6000$$

\begin{align} &= 12\% {\rm{ }}of{\rm{ }}6000 + 12\% {\rm{ }}of{\rm{ }}5500 + \ldots + 12\% {\rm{ }}of{\rm{ }}500\\ &= 12\% {\rm{ }}of\left( {6000 + 5500 + \ldots + 500} \right)\\ &= 12\% {\rm{ }}of\left( {500 + 1000 + \ldots + 6000} \right)\end{align}

Now, the series $$500 + 1000 + \ldots + 6000$$ forms an A.P. with $$a$$ and $$d$$ both equal to 500.

Let the number of terms of the A.P. be $$n$$

Therefore,

\begin{align} &\Rightarrow {a_n} = a + \left( {n - 1} \right)d\\ &\Rightarrow 6000 = 500 + \left( {n - 1} \right)\left( {500} \right)\\ &\Rightarrow 6000 - 500 = 500\left( {n - 1} \right)\\& \Rightarrow n - 1 = \frac{{5500}}{{500}}\\&\Rightarrow n = 11 + 1\\&\Rightarrow n = 12\end{align}

Hence,

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{12}} &= \frac{{12}}{2}\left[ {2\left( {500} \right) + \left( {12 - 1} \right)\left( {500} \right)} \right]\\ &= 6\left[ {1000 + 5500} \right]\\& = 6 \times 6500\\ &= 39000\end{align}

Therefore, total interest to be paid is

\begin{align}12\% {\rm{ }}of\left( {500 + 1000 + \ldots + 6000} \right) &= 12\% {\rm{ }}of{\rm{ }}39000\\ &= \frac{{12}}{{100}} \times 39000\\ &= 4680\end{align}

Now, total cost of tractor in $$₹$$ $$= 12000 + 4680 = 16680$$

Thus, the tractor will cost him $$₹$$ $$16680$$.

## Chapter 9 Ex.9.ME Question 28

Shamshad Ali buys a scooter for $$₹$$$$22000$$. He pays $$₹$$$$4000$$cash and agrees to pay the balance in annual instalment of $$₹$$$$1000$$ plus $$10\%$$ interest on the unpaid amount. How much will the scooter cost him?

### Solution

It is given that Shamshad Ali buys a scooter for $$₹$$$$22000$$ and pays $$₹$$$$4000$$ in cash.

Hence, unpaid amount in $$₹$$ $$= 2{\rm{2}}000 - 4000 = 18000$$

According to the given condition, the interest paid annually is

$$\left( {10\% {\rm{ }}of{\rm{ 18}}000} \right),\left( {10\% {\rm{ }}of{\rm{ 1700}}0} \right),\left( {10\% {\rm{ }}of{\rm{ 160}}00} \right), \ldots \left( {10\% {\rm{ }}of{\rm{ 10}}00} \right)$$

Thus, total interest to be paid

\begin{align} &= \left( {10\% {\rm{ of}}{\rm{ 18}}000} \right) + \left( {10\% {\rm{ of}}{\rm{ 1700}}0} \right) + \ldots + \left( {10\% {\rm{ }}of{\rm{ 10}}00} \right)\\ &= 10\% {\rm{ }}of\left( {18000 + 17000 + \ldots + 1000} \right)\\ &= 10\% {\rm{ }}of\left( {1000 + 2000 + \ldots + 18000} \right)\end{align}

Now, the series $$1000 + 2000 + \ldots + 18000$$forms an A.P. with $$a$$ and $$d$$ both equal to 1000.

Let the number of terms of the A.P. be $$n$$

Therefore,

\begin{align} &\Rightarrow {a_n} = a + \left( {n - 1} \right)d\\ &\Rightarrow 18000 = 1000 + \left( {n - 1} \right)\left( {1000} \right)\\& \Rightarrow 18000 - 1000 = 1000\left( {n - 1} \right)\\ &\Rightarrow n - 1 = \frac{{17000}}{{1000}}\\ &\Rightarrow n = 17 + 1\\ &\Rightarrow n = 18\end{align}

Hence,

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{18}} &= \frac{{18}}{2}\left[ {2\left( {1000} \right) + \left( {18 - 1} \right)\left( {1000} \right)} \right]\\ &= 9\left[ {2000 + 17000} \right]\\ &= 9 \times 19000\\ &= 171000\end{align}

Therefore, total interest to be paid is

\begin{align}10\% \,{\rm{ of}}\left( {1000 + 2000 + \ldots + 18000} \right) &= 10\% {\rm{ }}of{\rm{ 171}}000\\ &= \frac{{10}}{{100}} \times 171000\\ &= 17100\end{align}

Now, total cost of scooter in $$₹$$ $$= 22000 + 17100 = 39100$$

Thus, the scooter will cost him $$₹$$$$39100$$.

## Chapter 9 Ex.9.ME Question 29

A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs $$50$$ paise to mail one letter, find the amount spent on the postage when $${8^{th}}$$ set of letter is mailed.

### Solution

The number of letters mailed forms a G.P: $$4,{4^2}, \ldots ,{4^8}$$

Here, $$a = 4,r = 4,n = 8$$

Hence,

\begin{align}{S_n}& = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\\{S_8} &= \frac{{4\left( {{4^8} - 1} \right)}}{{4 - 1}}\\ &= \frac{{4\left( {65536 - 1} \right)}}{3}\\ &= \frac{{4 \times 65535}}{3}\\ &= 4 \times 21845\\ &= 87380\end{align}

It is given that the cost to mail one letter is $$50$$ paisa

Therefore, cost in $$₹$$ for mailing $$87380$$ letters $$= \frac{{50}}{{100}} \times 87380 = 43690$$

Thus, $$₹$$$$43690$$spent when on the postage when $${8^{th}}$$ set of letter is mailed.

## Chapter 9 Ex.9.ME Question 30

A man deposited $$₹$$$$10000$$ in a bank at the rate of $$5\%$$ simple interest annually. Find the amount in the $${15^{th}}$$ year since he deposited the amount and also calculate the total amount after 20 years.

### Solution

It is given that the man deposited $$₹$$$$10000$$in a bank at the rate of $$5\%$$ simple interest annually

Hence, annual interest received in $$₹$$ $$= \frac{5}{{100}} \times 10000 = 500$$

Amount in $$₹$$ at the end of first year $$= 10000 + 500 = 10500$$

Amount in $$₹$$ at the end of second year $$= 10500 + 500 = 11000$$

Amount in $$₹$$ at the end of third year $$= 11000 + 500 = 11500$$

Therefore, $$10000,10500,11000, \ldots$$ is an A.P.

Hence, the amount in the $$15^{th}$$ year

\begin{align}{a_{15}}& = 10000 + \left( {15 - 1} \right)\left( {500} \right)\\ &= 10000 + 7000\\ &= 17000\end{align}

Amount after $$20$$ years

\begin{align}{a_{21}} &= 10000 + \left( {21 - 1} \right)\left( {500} \right)\\ &= 10000 + 10000\\ &= 20000\end{align}

Thus, the amount in $$15^{th}$$ year is $$₹$$$$17000$$ and after 20 years amount is $$₹$$$$20000$$.

## Chapter 9 Ex.9.ME Question 31

A manufacturer reckons that the value of a machine, which costs him $$₹$$$$15625$$, will depreciate each year by $${\rm{2}}0\%$$. Find the estimated value at the end of $$5$$ years.

### Solution

Cost of machine is $$₹$$$$15625$$

Machine depreciates by $${\rm{2}}0\%$$ every year.

Therefore,

its value after every year is $$80%$$ of the original cost i.e., $$\frac{4}{5}$$of the original cost.

Value at end of $$5$$ years $$= 15625 \times \underbrace {\frac{4}{5} \times \frac{4}{5} \times \frac{4}{5} \times \ldots \times \frac{4}{5}}_{5{\rm{ }}times} = 15625 \times {\left( {\frac{4}{5}} \right)^5} = 5 \times 1024 = 5120$$

Thus, value of the machine at the end of $$5$$ years is $$₹$$ $$5120$$.

## Chapter 9 Ex.9.ME Question 32

$$150$$ workers were engaged to finish a job in a certain number of days. $$4$$ workers dropped out on second day, $$4$$ more workers dropped out on third day and so on. It took $$8$$ more days to finish the work. Find the number of days in which the work was completed.

### Solution

Let $$x$$ be the number of days in which $$150$$ workers finish the work.

According to the given information,

$150x = 150 + 146 + 142 + \ldots \left( {x + 8} \right){\rm{ terms}}$

Here, $$150+146+142+\ldots \left( x+8 \right)\text{ terms}$$ forms an A.P. with $$a=146,d=-4,n=\left( x+8 \right)$$

Therefore,

\begin{align}& \Rightarrow 150x = \frac{{\left( {x + 8} \right)}}{2}\left[ {2\left( {150} \right) + \left( {x + 8 - 1)( - 4} \right)} \right]\\ &\Rightarrow 150x = \left( {x + 8} \right)\left[ {150 + \left( {x + 7} \right)\left( { - 2} \right)} \right]\\ &\Rightarrow 150x = \left( {x + 8} \right)\left( {150 - 2x - 14} \right)\\ &\Rightarrow 150x = \left( {x + 8} \right)\left( {136 - 2x} \right)\\ &\Rightarrow 75x = \left( {x + 8} \right)\left( {68 - x} \right)\\ &\Rightarrow 75x = 68x - {x^2} + 544 - 8x\\ &\Rightarrow {x^2} + 15x - 544 = 0\\& \Rightarrow {x^2} + 32x - 17x - 544 = 0\\& \Rightarrow x\left( {x + 32} \right) - 17\left( {x + 32} \right) = 0\\ &\Rightarrow \left( {x - 17} \right)\left( {x + 32} \right) = 0\\ &\Rightarrow x = 17, - 32\end{align}

However, $$x$$ cannot be negative, hence,$$x = 17$$

Therefore, the number of days in which the work was completed is $$\left( {17 + 8} \right) = 25$$ days.

Thus, the work was completed in $$25$$ days.

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