Question 14 Marks
If the fourth term and the seventh term are $\frac{3}{4}$ and $\frac{3}{32}$ respectively for a $G.P.,$ find its tenth term.
AnswerLet $a$ and $r$ be the first term and common ratio respectively of the given $G.P.$
We are given $\mathrm{T}_{4}=\frac{3}{4}$ and $\mathrm{T}_{7}=\frac{3}{32}$
Hence, $\frac{T_{7}}{T_{4}}=\frac{a r^{6}}{a r^{3}}=r^{3}$ and $\frac{T_{7}}{T_{4}}=\frac{\frac{3}{32}}{\frac{3}{4}}=\frac{3}{32} \times \frac{4}{3}=\frac{1}{8}$
$ \begin{aligned} &\therefore \quad r^{3}=\frac{1}{8}=\left(\frac{1}{2}\right)^{3} \\ &\therefore \quad r=\frac{1}{2} \end{aligned} $
Now putting $r=\frac{1}{2}$ in $\mathrm{T}_{4}=a r^{3}=\frac{3}{4}$, we get
$ \begin{array}{ll} \therefore & a\left(\frac{1}{2}\right)^{3}=\frac{3}{4} \\ \therefore & a \times \frac{1}{8}=\frac{3}{4} \\ \therefore & a=\frac{3}{4} \times 8 \\ \therefore & a=6 \end{array} $
We now have to find the tenth term, so $n=10$.
Putting values of $a, r$ and $n$ in the general term $\mathrm{T}_{n}=a r^{n-1}$, we get
$\begin{aligned}\mathrm{T}_{10} &=6 \times\left(\frac{1}{2}\right)^{10-1} \\&=6 \times\left(\frac{1}{2}\right)^{9} \\&=6 \times \frac{1}{512} \\&=\frac{3}{256}\end{aligned}$
Hence, the tenth term of the $G. P.$ is $\frac{3}{256}$
View full question & answer→Question 24 Marks
The sum and the product of the three consecutive terms of a $GP$. are $9.5$ and $27$ respectively. Find the three terms of the $GP.$
Answer Let us assume three consecutive terms of G.P. as $\frac{a}{r}, a, a r$
Here the product of the terms $=27$$\therefore \frac{a}{r} \times a \times a r=27$
$\therefore a^{3}=27$
$\therefore a=3$
Now the sum of the terms $=9.5$
$\therefore \frac{a}{r}+a+a r=9.5$
$\therefore a\left(\frac{1}{r}+1+r\right)=9.5$
$\therefore 3\left(\frac{1+r+r^{2}}{r}\right)=9.5$
$\therefore 3+3 r+3 r^{2}=9.5 r$
$\therefore 3 r^{2}-6.5 r+3=0$
Multiplying by $2$ on both the sides, we get
$ \therefore 6 r^{2}-13 r+6=0$
$ \therefore \quad(2 r-3)(3 r-2)=0$
$ \therefore r=\frac{3}{2} \text { or } r=\frac{2}{3}$
Now, if we take $a=3$ and $r=\frac{3}{2}$
then three consecutive terms will be $\frac{3}{\left(\frac{3}{2}\right)}=2,3,3 \times \frac{3}{2}=\frac{9}{2}$
Hence, three consecutive terms are $2,3, \frac{9}{2}$
Now, if we take $a=3$ and $r=\frac{2}{3}$
then three consecutive terms will be $\frac{3}{\frac{2}{3}}=\frac{9}{2}, 3,3 \times \frac{2}{3}=2$
Hence, three consecutive terms are $\frac{9}{2}, 3,2$.
View full question & answer→Question 34 Marks
The sum and the product of the three consecutive terms of a $GP.$ are $26$ and $216$ respectively. Find the three terms of the $GP.$
AnswerLet us assume three consecutive terms of $G.P.$ as $\frac{a}{r}, a, a r$
Here, the product of the terms $=216$
$ \therefore \frac{a}{r} \times a \times a r =216$
$\therefore a^{3} =216$
$\therefore a =6$
Now, the sum of the terms $=26$
$ \therefore \frac{a}{r}+a+a r=26$
$ \therefore a\left(\frac{1}{r}+1+r\right)=26$
$ \therefore 6\left(\frac{1+r+r^{2}}{r}\right)=26 \quad(\because a=6)$
$ \therefore 3\left(1+r+r^{2}\right)=13 r$
$ \therefore 3+3 r+3 r^{2}=13 r$
$ \therefore 3 r^{2}-10 r+3=0$
$ \therefore \quad(r-3)(3 r-1)=0$
$ \therefore r=3 \text { or } r=\frac{1}{3}$
Now, if we take $a=6$ and $r=3$
then three consecutive terms will be $\frac{6}{3}=2,6,6 \times 3=18$.
Hence, three consecutive terms are $2,6,18$.
Now, if we take $a=6$ and $r=\frac{1}{3}$
then three consecutive terms will be $\frac{6}{1 / 3}=18,6,6 \times \frac{1}{3}=2$
Hence, three consecutive terms are $18,6$ and $2 .$
View full question & answer→Question 44 Marks
A person wants to donate ? $2,42,000$ in five months such that every month he donates one-third of the amount he donated in the previous month. Find the amount he donated in the first month.
Answer The person wants to donate $₹ 242000$ in five months
i.e. $\mathrm{S}_{5}=242000$ and $n=5$.
Every month, he donates one-third of the amount he donated in the previous month, so the common ratio $r=\frac{1}{3}$.
We want to find the amount donated in the first month
i.e. $a$.Putting the values of $a, r$ and $n$ in $\mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{(r-1)}$,
we get,$\mathbf{S}_{5}=\frac{a\left[\left(\frac{1}{3}\right)^{5}-1\right]}{\left(\frac{1}{3}-1\right)}$
$\therefore 242000=\frac{a\left(\frac{1}{243}-1\right)}{\left(-\frac{2}{3}\right)}$
$\therefore 242000=\frac{a\left(-\frac{242}{243}\right)}{\left(-\frac{2}{3}\right)}$
$\therefore 242000-a \times \big( \frac{-242}{ 243} \big) \times \big(\frac{3}{-2}\big)$
$\therefore 242000=a \times \frac{121}{81}$
$\therefore a=\frac{242000 \times 81}{121}$
$\therefore a=162000$
Thus, the amount donated in the first month is $₹ 1,62,000$.
View full question & answer→Question 54 Marks
Find the minimum value of $n$ such that the sum of the first $n$ terms of a $G$. P. $1,3,3^{2}, 3^{3}, \ldots$ is greater than or equal to $3000 .$
AnswerHere, the first term $a=1$ and the common ratio $r=\frac{3}{1}=3$.
Sum of the first $n$ terms should be greater than or equal to $3000$
i.e. $\mathrm{S}_{n} \geq 3000$.Putting the values of $a, r$ and $n$ in $\mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{(r-1)}$,
we get,$\mathrm{S}_{n} =\frac{1\left(3^{n}-1\right)}{(3-1)}$
$ =\frac{3^{n}-1}{2}$Since, $\mathrm{S}_{n} \geq 3000$,
we have$ \frac{3^{n}-1}{2} \geq 3000$
$\therefore 3^{n}-1 \geq 6000$
$\therefore 3^{n} \geq 6001$
We now tabulate values of $3^{\text {n }}$ for different positive values of $n$ as shown in the following table and
we shall take minimum value of $n$ for which $3^{n} \geq 6001$.
We can see from the table that when we take $n=8,9, \ldots, 3^{n}$ exceeds $6001$
Thus, the minimum value of $n$ for which $3^{n}$ is greater than or equal to $6001$ is $8$ .
$\therefore n=8$ View full question & answer→Question 64 Marks
Find the maximum value of n such that the sum of the first $11$ terms of a $G. P. 2, 4, 8, 16 ......$ does not exceed $5000.$
AnswerHere, the first term $a=2$ and the common ratio $r=\frac{4}{2}=2$.
Sum of the first $n$ terms should not exceed 5000 i.e. $\mathrm{S}_{n} \leq 5000$.
Putting the values of $a, r$ and $n$ in $\mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{(r-1)}$,
we get$\mathrm{S}_{n} =\frac{2\left(2^{n}-1\right)}{(2-1)}$
$ =2\left(2^{n}-1\right)$
Since $\mathrm{S}_{n} \leq 5000$, we have$ 2\left(2^{n}-1\right) \leq 5000$
$\therefore 2^{n}-1 \leq 2500$
$\therefore 2^{n} \leq 2501$
We now tabulate values of $2^{n}$ for different positive values of $n$ as shown in
the following table and we shall take the maximum value of $n$ for which $2^{\circ} \leq 2501$
We can see from the table that for $n=10,2^{n}=1024$,
for$n=11,2^{n}=2048$ and for $n=12$. $2^{n}=4096$
which exceeds $2501$ so the maximum value of $n$
for which $2^{n}$ does not exceed $2501$ is $11 $.
$\therefore n=11$ View full question & answer→Question 74 Marks
A construction company offers a scheme on a flat to attract customers. In this scheme, customer has to pay $Rs 10,000$ as the first installment and has to pay double the amount of the preceeding installment in the subsequent annual installments. What is the total amount that the customer has to pay upto $10$ installments ?
AnswerHere, the first installment $a=R s 10,000$
Yearly installment is double the amount of the preceeding installment.
$\therefore r=2$
$10$ installments are to be paid.
$\therefore n =10$
The total amount that the customer has to Pay up to $10$ installments $=S_n$
Now, $S_n=\frac{a\left[r^n-1\right]}{r-1}$
Putting $n=10 ; a=10,000 ; r=2$,
$S_{10}=\frac{10000\left[(2)^{10}-1\right]}{r-1}$
$=10,000(1024-1)$
$=10,000 \times 1023$
$=Rs \ 1,02,30,000$
Hence, the customer has to pay Rs $1,02,30,000$
View full question & answer→Question 84 Marks
The sum and the Product of the three consecutive terms of $G.P.$ are $6$ and $-64$ respectively. Find the three terms of the $G.P.$
AnswerSuppose, three consecutive terms of $G.P.$ are $\frac{ a }{ r ^{\prime}} a , ar$,
The sum of the three consecutive terms $=6$
$\therefore \frac{a}{r}+a+a r=6$
$\therefore \frac{a+a r+a r^2}{r}=6$
$\therefore a+a r+a r^2=6 r.$
The product of the three consecutive terms $=-64$
$\therefore \frac{a}{r} \times a \times a r=-64$
$\therefore a^3=-64=(-4)^3$
$\therefore a=-4$
Putting $a =-4$ in the result $(1),$
$-4-4 r-4 r^2=6 r$
$\therefore 4 r^2+4 r+6 r+4=0$
$\therefore 4 r^2+10 r+4=0$
$\therefore 2 r^2+5 r+2=0$
$\therefore 2 r^2+4 r+r+2=0$
$\therefore 2 r(r+2)+1(r+2)=0$
$\therefore(r+2)(2 r+1)=0$
$\therefore r+2=0$ OR $2 r+1=0$
$\therefore r=-2$ OR $r=-\frac{1}{2}$
Putting $a =-4$ and $r =-2$ in $\frac{ a }{ r ^{\prime}} a ; \frac{ a }{ r ^{\prime}}$
Three consecutive terms :
$\frac{-4}{-2^{\prime}}-4,(-4)(-2)$
$ \Longrightarrow 2,-4,8.$
Putting $a =-4$ and $r =-\frac{1}{2}$ in $\frac{ a }{ r ^{\prime}}$ a; $\frac{ a }{ r ^{\prime}}$
Three consecutive terms :
$\frac{-4}{-\frac{1}{2}},-4,(-4)\left(-\frac{1}{2}\right) \Rightarrow 8,-4,2.$
Hence, three consecutive terms are $2,-4,8$ OR $8,-4,2$.
View full question & answer→Question 94 Marks
For a geometric progression, the ratio of sum of the fifth and the third term to the difference of the fifth and the third term is $5 : 3.$ Find $r.$
AnswerHere, $\left(T_5+T_3\right):\left(T_5-T_3\right)=5: 3$
Now, $T_n=a \cdot r^{n-1}$
$\therefore T_5=a r^4$ and $T_3=a r^2$
$\left(T_5+T_3\right):\left(T_5-T_3=5: 3\right.$
$\therefore \frac{a r^4+a r^2}{a r^4-a r^2}=\frac{5}{3}$
$\therefore \frac{a r^2\left(r^2+1\right)}{a r^2\left(r^2-1\right)}=\frac{5}{3}$
$\therefore 3 r^2+3=5 r^2-5$
$\therefore 3+5=5 r^2-3 r^2$
$\therefore 8=2 r^2$
$\therefore r^2=\frac{8}{2}=4$
$\therefore r= \pm 2$
View full question & answer→Question 104 Marks
for a geometric progression, $S _4=10 S_2$. Find the common ratio.
Answer$S_4=10 S_2$
$S_n=\frac{a\left[r^n-1\right]}{r-1}$
$\therefore S_4=\frac{a\left[r^4-1\right]}{r-1} \text { and } S_2=\frac{a\left\lfloor r^2-1\right]}{r-1}$
Now, $S_4=10 S_2$
$\frac{a\left[r^4-1\right]}{r-1}=10 \times \frac{a\left[r^2-1\right]}{r-1}$
$\therefore r^4-1=10\left(r^2-1\right)$
$\therefore\left(r^2-1\right)\left(r^2+1\right)=10\left(r^2-1\right)$
$\therefore\left(r^2+1\right)=10$
$\therefore r^2=10-1=9$
$\therefore r= \pm 3$
View full question & answer→Question 114 Marks
The sum of the first terms of the $G.P. y.$ $\frac{ y }{ 3 }, \frac{ y }{9}, \ldots$ (where, $y>0$ ) is $121.$ Find $y$.
AnswerHere, $a=y ; r=\frac{\frac{y}{2}}{y}=\frac{1}{3^{\prime}} S_5=121$
$S_n=\frac{a\left[1-r^n\right]}{1-r}(\because r < 1)$
Putting $n=5 ; a=y ; r=\frac{1}{3}$.
$S_5=\frac{y\left[1-\left(\frac{1}{8}\right)^5\right]}{1-\frac{1}{3}}$
$\therefore 121=\frac{y\left[1-\frac{1}{241}\right]}{\frac{2}{3}}$
$\therefore 121 \times \frac{2}{3}= y \left[\frac{243-1]}{243}\right.$
$\therefore \frac{243}{3}= y \left(\frac{242}{243}\right)$
$\therefore Y=\frac{242}{3} \times \frac{243}{242}$
$\therefore Y=81$
View full question & answer→Question 124 Marks
Find the minimum value of $n$ such that the sum of the first $n$ terms of a $G.P.$ $1,2,2^2, 2^3, \ldots$ is greater than or equal to $2000.$
AnswerHere, $a=1 ; r=2$, The sum of first $n$ terms is greater than or equal to $2000$ ,
i.e., $S_n \geq 2000$.
Now, $S_n=\frac{a\left[ r ^{ n }-1\right]}{ r -1}$
Putting $a=1 ; r=2$,
$\left.S_n=\frac{1\left[2^n-1\right]}{2-1}=2^n-1 \right\rvert\,$
$S_n \geq 2000$
$\therefore 2^n-1 \geq 2000$
$\therefore 2^n \geq 2000+1$
$\therefore 2^n \geq 2001$
We now tabulate values of $2^n$ for different positive values of $n$ as shown in the following table and we shall take the minimum value of $n$ for which $2^n \geq 2001$.
| $n$ |
$8$ |
$9$ |
$10$ |
$11$ |
$12$ |
| $2^n$ |
$256$ |
$512$ |
$1024$ |
$2048$ |
$4096$ |
From the table we can see that for $n=112^n=2048$ which is greater than $2001.$
$\therefore$ For $n=11,2^n \leq 2001$.
Hence, $n=11$ is minimum value of $n$. View full question & answer→Question 134 Marks
Find the maximum value of $n$ such that the sum of the first $n$ terms of a $G.P. 1,3,3^2, 3^3$ ..... does not exceed $365.$
AnswerHere, $a=1, r=\frac{3}{1}=3$; the sum of the first $n$ terms does not exceed $365$ ,
i.e., $S_n \leq 365$.
Now, $S _{ n }=\frac{ a \left[ r ^{ n }-1\right]}{ r -1}$
Putting $a =1 ; r =3$,
$S_n=\frac{1\left[3^{ n }-1\right]}{3-1}$
$S_n=\frac{\left[3^{ n }-1\right]}{2}$
Now, $S _{ n } \leq 365$
$\therefore \frac{\left[3^n-1\right]}{2} \leq 365$
$\therefore 3^n-1 \leq 365 \times 2$
$\therefore 3^n \leq 730+1$
$\therefore 3^n \leq 731$
We know tabulate values of $3^n$ for different positive values of $n$ as shown in the following table ans we shall take the maximum value of $n$ for which $3^n \leq 731$.
| $n$ |
$5$ |
$6$ |
$7$ |
| $3^n$ |
$243$ |
$729$ |
$2187$ |
For $n=7,3^n=2187$ which is more than $731.$
$\therefore n=6$ is the maximum value of $n$, for which $3^{n=} \leq 731$. View full question & answer→Question 144 Marks
A car depreciates at the rate of $10%$ every year. If the cost price of the car is $Rs 5,00,000$ what will be the value of the car after $6$ years ?
AnswerThe cost price of the car is$ Rs 5,00,000$.
$\therefore a=5,00,000$
The value of the car depreciates at the rate of $10 \%$ every year.
$\therefore r =\frac{100-10}{100}=\frac{90}{100}=0.9$
The value of the car after $6$ years $=T_7$
Now, $T_n=a . r^{n-1}$
Putting $n=7 ; a=5,00,000 ; r=0.9$,
$\therefore T_7=5,00,000(0.9)^{7-1}$
$=5,00,000(0.9)^6$
$=5,00,000(0.531441)$
$=Rs 2,65,720.50$
Hence, the value of the car after $6$ years will be Rs $2,65,720.50$.
View full question & answer→Question 154 Marks
Population of a village is $5000.$ Population increases at the rate of $2\%$ every year.What will be the population of the village after $10$ years ?
AnswerPopulation of a village is $5000.$
$\therefore a=5000$
Population increases at the rate of $2 \%$ every year.
$102$
$\therefore r =\frac{102}{100}=1.02$
The population after 10 years $=T_{11}$
Now. $T_n=a . r^{n-1}$
Putting $n=11 ; a=5000 ; r=1.02$,
$T_{11}=5000(1.02)^{11-1}$
$=5000(1.02)^{10}$
$=5000(1.219)$
$=6095$
$(1.02)^{10}$
$=A L[10 \log 1.02]$
$=A L[10 \times 0.0086]$
$=A L[0.086]=1.219$
Hence, the population of the village after $10$ years will be $6095.$
View full question & answer→Question 164 Marks
A banker counts $128$ notes in the first minute and there after he counts half the number of notes he counted in the previous minute. How many notes he would count in five minutes ?
AnswerIn the first minute a banker counts $128$ notes.
$\therefore a =128$
There after, in every minute he counts half the number of notes he counted in the previous minute.
$\therefore r =\frac{1}{2}$
Total number of notes counted in minutes $= S _5$
Now, $S_n=\frac{a\left[1-r^{ n }\right]}{1-r}$
Putting $n=5 ; a=128 ; r=\frac{1}{2}$.
$S_5=\frac{128\left[1-\left(\left(\frac{1}{2}\right)^3\right]\right.}{1-\frac{1}{2}}$
$=\frac{128\left[1-\frac{1}{22}\right]}{\frac{1}{2}}$
$=128 \times 2\left[\frac{32-1}{32}\right]$
$=4 \times 2(31)=248$ Notes
View full question & answer→Question 174 Marks
If the three positive numbers $k + 4, 4k -2$ and $7k + 1$ are in $G.P.$, find $K.$
Answer$k +4,4 k -2,7 k +1$ are in $G.P.$
$\therefore \frac{4 k -2}{ k +4}=\frac{7 k +1}{4 k -2}$
$\therefore(4 k -2)(4 k -2)=(7 k +1)( k +4)$
$\therefore 16 k ^2-16 k +4=7 k ^2+28 k + k +4$
$\therefore 16 k ^2-7 k ^2-16 k -28 k - k +4-4=0$
$\therefore 9 k ^2-45 k =0$
$\therefore 9 k ( k -5)=0$
$\therefore 9 k =0 \text { OR } k -5=0$
$\therefore k =0 \text { OR } k =5$
Numbers are positive.
Hence, $k=5$
View full question & answer→Question 184 Marks
For a $G.P.$ if $T_{3}=4$ and $T_{5}=16$, find its common ratio and the first term.
View full question & answer→Question 194 Marks
A machine is purchased in $Rs. 10,600.$ The depreciation on the machine is calculated at the rate of $10\%$ per annum. What will be the estimated value of the machine at the end of four years?
View full question & answer→Question 204 Marks
A $TV$ passes through three stages before it reaches to a customer from the producer. At every stage the cost of $TV$ is increased by $10 \%$. If its production cost is $Rs. 12000$, what price will the customer pay?
View full question & answer→Question 214 Marks
A person has to repay the debt of $Rs. 6350$ in seven instalments. He repays the first instalment of large amount. After then he pays each instalment half of the previous instalment Find the amount of first and last instalment.
AnswerFirst instalment: $Rs \ 3200$, Last instalment: $Rs \ 50$
View full question & answer→Question 224 Marks
If $S_{n}=8\left[\left(\frac{3}{2}\right)^{n}-1\right]$, find $T_{n+1}$ and hence the value of $T_{2}+T_{3}$.
Answer$T_{n+1}=4\left(\frac{3}{2}\right)^{n}, 15$
View full question & answer→Question 234 Marks
If $S_{n}=\frac{1}{54}\left[3^{n}-1\right]$, find $T_{n+1}$ and hence Write the $G.P.$
Answer$\overline{T_{n+1}}={ }_{27}^{1} \cdot 3^{n} ;=3^{n-3} ;{ }_{9}^{1},{ }_{3}^{1}, 1, \ldots \ldots$
View full question & answer→Question 244 Marks
In a housing scheme a customer has to pay the cost of a house in $20$ annual instalments. He has to pay $Rs. 1000$ in the first instalment and there after he has to pay double amount of the previous instalment. How much amount will he pay in $10th$ yearly?
View full question & answer→Question 254 Marks
In a state at the rate of $3 \%$ the number of children increases. If in $2012$, the number of children was $12000$, what will be the number of children in the beginning of the year $2017?$ Also find the number of children at the end of year $2017.$
Answer$13506$ children, $13911$ children
View full question & answer→Question 264 Marks
A water tank is full of $2000$ litres of water. Due to leakage everyday ${ }_{4}^{1}$ th of tank is being empty. On inspection it was found that only $125$ litres of water remained in the tank. Find after how many days will the inspection carried out.
View full question & answer→Question 274 Marks
A person bays a maruti car at $Rs. 4,50,000.$ If $10 \%$ depreciation is available on the cost of the car per year, what will be the cost of the car after $4$ years?
View full question & answer→Question 284 Marks
A person donates $Rs. 1000$ in January, $2017$ and $Rs. 2000$ in February, $2017$. Thus every month he donates double the amount than the previous month. How much can he donate in January $2017?$
View full question & answer→Question 294 Marks
The population of a city was $5000$ in $2012$. If the growth rate is $2 \%$ per annum, what will be the population of the city in $2017?$
View full question & answer→Question 304 Marks
A producer sell his product to wholesaler, thereafter he sells to retailer and retailer to customer. At every stage of sell the production cost increase by $10 \%$. If a customer pays $6655$ for the product, find the production cost of the producer.
View full question & answer→