Question 13 Marks
The first term and the product of the first three terms of a $G.P.$ are $3$ and $216$ respectively. Find the $7th$ term of the $G.P$.
AnswerThe first tern $a=3$.
Now, putting $n=1, n=2$ and $n=3$ in the general term $\mathrm{T}_n=a r^{n-1}$,
we get $\mathrm{T}_1=a, \mathrm{~T}_2=a r$ and $\mathrm{T}_3=a r^2$
It is given that product of first three terms is $216$ .
i.e. $\mathrm{T}_1 \times \mathrm{T}_2 \times \mathrm{T}_3=216$
Now, $T_1 \times T_2 \times T_3=216$
$\therefore a \times a r \times a r^2=216$
$\therefore a^3 \times r^3=216$
$\therefore 3^3 \times r^3=216 (\because a=3)$
$\therefore 27 \times r^3=216$
$\therefore r^3=\frac{216}{27}=8$
$\therefore r=2$
The $7th$ term of the $G.P.$ is $\mathrm{T}_{\mathrm{T}}=a r^5$
$\therefore \mathrm{T}_{\Psi} =3 \times(2)^6$
$ =3 \times 64$
$ =192$
Hence, the $7th$ term of the $\mathrm{G} P$ is $192$ .
View full question & answer→Question 23 Marks
If for a $G.P.$, $\mathrm{T}_{n}=80, \mathrm{~S}_{n}=157.5$ and $r=2$, find $a$ and $n$.
AnswerHere, values of $\mathrm{T}_n$ and $\mathrm{S}_n$ are given, so we shall use the formula
$\mathrm{S}_n=\frac{r T_m-a}{r-1}
$Putting the values of $T_n, r$ and $S_n$ in $S_e=\frac{r T_n-a}{r-1}$
$\therefore 157.5=\frac{2 \times 80-a}{2-1}$
$\therefore 157.5=160-a$
$\therefore a=160-157.5$
$\therefore a=2.5$
$\text { Now } T_s=a r^{n-1}$
$\therefore 80=2.5 \times(2)^{-1}$
$\therefore 2^{n-1}=32$
$\therefore 2^{-1}=2^5$
Equating the powers on both the sides, we get
$n-1=5$
$\therefore n n$
Thus, $a=2.5$ and $n=6$
View full question & answer→Question 33 Marks
If $15, x, 240, y$ are in $GR$, find the values of $x$ and $y$.
Answer
Here, $15, x, 240, y$ are in $G. P.$ so $\frac{T_{2}}{T_{1}}=\frac{T_{3}}{T_{2}}=\frac{T_{4}}{T_{3}}=$ common ratio $(r)$ Let us take $\frac{T_{2}}{T_{1}}=\frac{T_{3}}{T_{2}}$,
$\therefore \frac{x}{15} =\frac{240}{x}$
$\therefore x^{2} =15 \times 240$
$ =3600$
$\therefore x =\pm 60$
$\therefore x =60 \text { or }-60$
Also, $\frac{T_{3}}{T_{2}}=\frac{T_{4}}{T_{3}}$
$\therefore \frac{240}{x} =\frac{y}{240}$
$\therefore x y =240 \times 240$
$\therefore x y =57600$
Now, if we take $x=60$ then y= 960
and if we take $x=-60 then y=-960$
View full question & answer→Question 43 Marks
A person gives $Rs. 5$ to his son on $1\ st$ March, $Rs. 10$ on $2\ nd$ March, $Rs. 20$ on $3\ rd$ March and so on. Thus each day he gives double the amount than that of the previous day. Find the total amount he has given to his son upto $10\ th$ of March.
Answer
The person gives $₹ 5$ on $1\ st$ March, so the first term $a=5$.
Every day, the amount given to his son is twice the previous day's amount, so the common ratio $r=2$.
We want to find the total amount given by him to his son upto $10\ th$ March i.e. $\mathrm{S}_{10}$, so $n=10$.
Putting the values of $a, r$ and $n$ in $\mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{(r-1)}$, we get,
$\mathrm{S}_{10} =\frac{5\left(2^{10}-1\right)}{(2-1)}$
$ =\frac{5(1024-1)}{1}$
$ =5 \times 1023$
$ =5115$
Hence, the total amount the person gives to his son upto $10\ th$ March is $₹ 5115$.
View full question & answer→Question 53 Marks
The first term and the sum of the first five terms of the $GP.$ are equal to $1$ each. Find the common ratio of the $G. P$.
AnswerHere, the first term $a=1$ and sum of the first five terms is 1 i.e. $\mathrm{S}_{5}=1$.
Putting the values in $\mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{(r-1)}$,
we get$ \mathrm{S}_{5}=\frac{1\left(r^{5}-1\right)}{(r-1)}$
$\therefore 1=\frac{\left(r^{5}-1\right)}{(r-1)}$
$\therefore r^{5}-1=r-1$
$\therefore r^{5}=r$
$\therefore r^{4}=1$
$\therefore r=\pm 1$If we take $r=1$ and $a=1$ then the first five terms of the $G. P.$ will be $1,1,1,1,1$ whose sum is $5$ .
If we take $r=-1$ and $a=1$ then the first five terms of the $G. P.$ will be $1,-1,1,-1,1$ whose sum is $1$ .
But, in the data, the sum of the first five terms is given as $1$ , so the only possible value of $r$ is $-1$.
Thus the common ratio of the $G. P.$ is $-1$.
View full question & answer→Question 63 Marks
In a $G.P.$, the first term is $27$ and sum of the first three terms is $189$. Find the common ratio of the $GP$.
AnswerHere the first term $a=27$ and sum of the first three terms is $189$ i.e. $\mathrm{S}_{3}=189$.
Putting the values in $\mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{(r-1)}$, we get
$\begin{aligned}& \mathrm{S}_{3}=\frac{27\left(r^{3}-1\right)}{(r-1)} \\\therefore & 189=\frac{27(r-1)\left(r^{2}+r+1\right)}{(r-1)} \\\therefore & 189=27\left(r^{2}+r+1\right) \\\therefore & r^{2}+r+1=7 \\\therefore & r^{2}+r-6=0 \\\therefore &(r+3)(r-2)=0 \\\therefore & r=-3 \text { OR } r=2\end{aligned}$
Thus, the common ratio of the $G. P.$ is $-3$ or $2$ .
View full question & answer→Question 73 Marks
Sum of how many terms of a $GP. 800,400,200,$ is $1500$ ?
Answer
Here, the first term $a=800$ and the common ratio $r=\frac{400}{800}=0.5 .$
Sum of $n$ terms is $1500$ i.e. $\mathrm{S}_{n}=1500$
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{800\left[(0.5)^{n}-1\right]}{(0.5-1)}$
$\therefore 1500=\frac{800\left[(0.5)^{n}-1\right]}{(-0.5)}$
$\therefore (0.5)^{n}-1=\frac{1500 \times(-0.5)}{800}$
$\therefore (0.5)^{n}=1-0.9375$
$\therefore (0.5)^{n}=0.0625$
$ (0.5)^{n}=(0.5)^{4}$
Equating the powers on both the sides, we get
$n=4$
Thus, the sum of the first four terms of the $G. P.$ is $1500$ .
View full question & answer→Question 83 Marks
The sum of the first three terms of a $G.P.$ with common ratio $0.2$ is $0.496$. Find the first term of the $G.P.$
AnswerHere, the common ratio $r=0.2$ and sum of the first three term is =$\frac{a\left(r^{n}-1\right)}{(r-1)}$, we get
$ 0.496=\frac{a\left[0.2^{3}-1\right]}{0.2-1}$
$\therefore 0.496=\frac{a(0.008-1)}{(-0.8)}$
$\therefore 0.496=\frac{a(-0.992)}{(-0.8)}$
$\therefore a=\frac{0.496 \times(-0.8)}{(-0.992)}$
$\therefore a=0.4 $
Thus, the first term of the $G. P.$ is $0.4$.
View full question & answer→Question 93 Marks
Find the sum of the first six terms of a $GP. 8, 4, 2, ....$
AnswerHere, the first term $a=8$ and the common ratio $r=\frac{4}{8}=\frac{1}{2}$ Sum of the first six terms is required i.e. $n=6$.
Putting the values of $a, r$ and $n$ in $\mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{(r-1)}$, we get
$\therefore \quad \mathrm{S}_{6}=\frac{8\left[\left(\frac{1}{2}\right)^{6}-1\right]}{\left(\frac{1}{2}-1\right)}$
$\begin{aligned}&=\frac{8\left[\frac{1}{64}-1\right]}{\left(-\frac{1}{2}\right)} \\&=\frac{8\left[-\frac{63}{64}\right]}{\left(-\frac{1}{2}\right)} \\&=8 \times\left(-\frac{63}{64}\right) \times\left(\frac{2}{-1}\right) \\&=\frac{63}{4}\end{aligned}$
Thus, the sum of the first six terms of the $G. P.$ is $\frac{63}{4}$.
View full question & answer→Question 103 Marks
Find the sum of the first five terms of a $GP. 5, 15, 45, ..$
Answer
Here, the first term $a=5$ and the common ratio $r=\frac{15}{5}=3$.
Sum of the first five terms is required i.e. $n=5$.
Putting the values of $a, r$ and $n$ in $S_{n}=\frac{a\left(r^{n}-1\right)}{(r-1)}$, we get
$\mathrm{S}_{5} =\frac{5\left(3^{5}-1\right)}{(3-1)}$
$ =\frac{5(243-1)}{2}$
$ =\frac{5(242)}{2}$
$ =605$
Thus, the sum of the first five terms of the $G. P.$ is $605$ .
View full question & answer→Question 113 Marks
Government decides to $‘x$ the depreciation rate of a machine to $15 \%$ per year. If the purchase price of a machine is $Rs. 50,000$. Find the value of the machine after $7$ years.
Answer
Here purchase price of a machine is $₹ 50,000$ i.e. $a=50,000=\mathrm{T}_{1} .$
Machine depreciates at the rate of $15 \%$ every year i.e. $\mathrm{r}=0.85$
As the machine depreciates, its price decreases at the rate of $15 \% \therefore r=\frac{100-15}{100}=0.85$
Value of machine after the first year $=50,000 \times 0.85$
$=42,500=\mathrm{T}_{2}$
We need to find the value of machine after $7$ years, so $n=8$.
Putting the values of $a, r$ and $n$ in the general term $\mathrm{T}_{n}=a r^{n-1}$,
$T_{8} =50,000(0.85)^{8-1}$
$ =50,000(0.85)^{7}$
$ =16028.8544$
$ \approx 16028.85$
Thus, the value of machine after $7$ years will be ₹ $16,028.85$
View full question & answer→Question 123 Marks
Population of a city is $20$ lakhs. If the population increases at the rate of $3\%$ every year, find the population of the city after $6$ years.
Answer
Current population of the city is $20,00,000$ i.e. $a=2000000=\mathrm{T}_{1}$.
Population increases at the rate of $3 \%$ every year
i.e. $r=1.03$
Population after the first year $=2000000 \times 1.03=2060000=\mathrm{T}_{2}$.
We need to find the population after $6$ years, so $n=7$.
Putting the values of $a, r$ and $n$ in the general term $\mathrm{T}_{n}=a r^{n-1}$, we get
$\mathrm{T}_{7} =20,00000(1.03)^{7-1}$
$ =20,00000(1.03)^{6}$
$\therefore \mathrm{T}_{7} =2388104.5930$
Thus, population after $6$ years will be $23,88,105$.
View full question & answer→Question 133 Marks
A water tank of a capacity of $50,000$ litres is fully ‘lled with water. Every week, the water level reduces to half of the previous level due to leakage. What will be the level of water after five weeks ? (No new water is added in the tank)
AnswerA water tank of a capacity of $50,000$ litres is fully filled with water
$\therefore a=50000=\mathrm{T}_{1}$.
Every week, the water level reduces to half of the previous level
$\therefore$ common ratio $r=\frac{1}{2}$.
Water level after first week $=50000 \times \frac{1}{2}=25000=\mathrm{T}_{2}$.
We want the water level after five weeks, so $n=6$.
Putting the values of $a, r$ and $n$ in the general term $\mathrm{T}_{n}=a r^{n-1}$, we get
$\mathrm{T}_{6} =50,000\left(\frac{1}{2}\right)^{6-1}$
$ =50,000\left(\frac{1}{2}\right)^{5}$
$ =50,000 \times \frac{1}{32}=1562.5$
Thus the level of water after five weeks will be $1562.5$ litres.
View full question & answer→Question 143 Marks
If the third term of a $GR$ is the square of the ‘rst term and the fourth term is $243$, find the sequence.
AnswerLet $a$ and $\mathrm{r}$ be the first term and common ratio respectively of the given $G. P.$ Here, the third term is square of the first i.e. $\mathrm{T}_{3}=a^{2}$
$\therefore \mathrm{T}_{3}=a r^{2}=a^{2}$
$\therefore r^{2}=a$
Also, the fourth term is 243 i.e. $T_{4}=243$$ \therefore \mathrm{T}_{4}=a r^{3}=243$
$ \therefore r^{2} \times r^{3}=243\left(\because r^{2}=a\right)$
$ \therefore r^{5}=243$
$ \therefore r^{5}=3^{5}$
$ \therefore r=3$
$ \text { Now, } a=r^{2}$
$ \therefore a=3^{2}$
$ \therefore a=9$
Taking $a=9$ and $r=3$ the sequence is $9,27,81,243, \ldots$
View full question & answer→Question 153 Marks
Which term of the $GP. 0.008, 0.016, 0.032,$ is $4.096$ ?
AnswerHere, the first term $a=0.008$ and the common ratio $r=\frac{0.016}{0.008}=2$
Now, $\mathrm{T}_{n}=4.096$
$\therefore a r^{n-1}=4.096$
$\therefore 0.008 \times(2)^{n-1}=4.096$
$\therefore 2^{n-1}=\frac{4.096}{0.008}$
$\therefore 2^{n-1}=512$
$\therefore 2^{n-1}=2^{9}$
Equating the powers on both the sides, we get$ n-1=9$
$\therefore n=10$
Hence, $4.096$ is the $10\ th$ term of the given $G.P.$
View full question & answer→Question 163 Marks
For a given $G.P.$, if $a=1, r=3$ and $S_n=121$; then find $n$. $a=1, r=3 S_n=121, n=$ ?
Answer$\mathrm{S}_n=\frac{a\left[r^n-1\right]}{r-1}$
$\therefore 242+1=(3)^n$
$\therefore 121=\frac{1\left[(3)^n-1\right]}{3-1}$
$\therefore 243=(3)^n$
$\therefore 121=\frac{(3)^n-1}{2}$
$\therefore(3)^5=(3)^n$
$\therefore 121 \times 2=(3)^n-1$
$\therefore n=5$
View full question & answer→Question 173 Marks
For a gives $G.P$, if $a =10, r =0.1$ and $T _{ n }=0.01$; then find $n$ .
Answer$a=10, r=0.1, T_n=0.01, n=?$
$T_n=a . r^{n-1}$
$\therefore 0.01=10(0.1)^{n-1}$
$\therefore \frac{0.01}{10}=(0.1)^{n-1}$
$\therefore 0.001=(0.1)^{n-1}$
$\therefore(0.1)^3=(0.1)^{n-1}$
$\therefore 3=n-1$
$\therefore n=4$
View full question & answer→Question 183 Marks
For a geometric progression, $T_1{ }^2=T_2$ and $T_3=64$. Write the sequence.
AnswerHere, $T_1^2=T_2 ; T_3=64$
$T_n=a \cdot r^{n-1}$
$T_1=a$
$T_2=a r$
$T_3=a r^2$
$\left(T_1\right)^2=T_2$
$(a)^2=a r$
$a=r$
$T_3=a r^2$
$64=r \cdot r^2$
$64=r^3$
$(4)^3=r^3$
$r=4$
$a=r: a=4$
First term $=4$ and common ratio $r=4$
Hence, the sequence obtained is $4,16,64, \ldots$
View full question & answer→Question 193 Marks
The first term of a $G.P.$, is $10$ and $T_4=0.08$. find sum of first three terms.
AnswerHere, $a =10, T_4=0.08, S_3=?$
$T_{ n }= a \cdot r ^{ n -1}$
$\therefore T _4=10 r ^3$
$\therefore 0.08=10 r ^3$
$\therefore r ^3=\frac{0.08}{10}=0.008$
$\therefore r ^3=(0.2)^3$
$\therefore r =0.2$
$S_n=\frac{a\left[1- r ^{ n }\right]}{1- r }(\because r < 1)$
$S _{ n }=\frac{10\left[1-(0.2)^3\right]}{1-0.2}$
$=\frac{10[1-0.008]}{1-0.2}$
$=\frac{10[1-0.008]}{0.8}$
$=\frac{10 \times 0.992}{0.8}$
$=\frac{9.92}{0.8}=12.4$
View full question & answer→Question 203 Marks
For a given $G.P.$, if $T_2=9$ and $T_5=243$; then find $S_4$.
AnswerHere, $T_2=9 ; T_5=243 ; S_4=$ ?
$T_n=a . r^{n-1}$
$\therefore T_2=a . r^{2-1}$
$\therefore 9=a r..........(1)$
$T_5=a . r^{5-1}$
$\therefore 243=a r^4.........(2)$
Taking ratio of results $(2)$ and $(1)$,
$\frac{243}{9}=\frac{a r^4}{ a r }$
$\therefore 27=r^3$
$\therefore(3)^3=r^3$
$\therefore r=3$
Putting $r=3$ in the result $(1),$
$9=3 a \therefore a=3$
Now, $S_n=\frac{a\left[r^n-1\right]}{r-1}(\because r > 1)$
$\therefore S_4=\frac{3\left[(3)^4-1\right]}{3-1}=\frac{3\lfloor 81-1]}{2}=\frac{3 \times 80}{2}=120$
View full question & answer→Question 213 Marks
For a given $G.P.$, $a =4$ and $T _5=\frac{ 1 }{ 4 }$; then find $T _7$.
AnswerHere, $a =4 ; T _5=\frac{1}{4} ; T _7=$ ?
$T_n=a \cdot r^{n-1}$
$\therefore T_5=4(r)^{5-1}$
$\therefore \frac{1}{4}=4 r^4$
$\therefore \frac{1}{4} \times \frac{1}{4}=r^4$
$\therefore \frac{1}{16}=r^4$
$\therefore\left(\frac{1}{2}\right)^4=r^4$
$\therefore r=\frac{1}{2}$
$T_7=a \cdot r^{7-1}$
$\therefore T_7=4\left(\frac{1}{2}\right)^6$
$=4 \times \frac{1}{64}$
$=\frac{1}{16}$
View full question & answer→Question 223 Marks
Find $T _5$ and $S _4$ of the geometric progression if the first term is $\frac{27}{16}$ and common ratio is $\frac{2}{3}$.
AnswerHere, $a =\frac{27}{16^{\prime}} ; r =\frac{2}{3} ; T _5=$ ? $S _4=$ ?
$T_n=a \cdot r^{n-1}$
$\therefore T_5=\left(\frac{27}{16}\right)\left(\frac{2}{3}\right)^{5-1}=\left(\frac{27}{16}\right)\left(\frac{2}{3}\right)^4=\frac{27}{16} \times \frac{16}{81}=\frac{1}{3}$
$S_n=\frac{a\left[1-r^n\right]}{1-r}(\because r < 1)$
$\therefore S_4=\frac{\frac{27}{16}\left[1-\left(\frac{2}{3}\right)^4\right]}{1-\frac{2}{8}}=\frac{\frac{27}{16}\left[1-\frac{16}{81}\right]}{\frac{2}{3}}=\frac{27 \times 3}{16}\left[\frac{81-16}{81}\right]=\frac{81}{16} \times \frac{65}{81}=\frac{65}{16}$
$\therefore S_4=\frac{65}{16}$
View full question & answer→Question 233 Marks
Find the sum of required terms for the following sequence using series formula :
$(1)$ $4,16,64, \ldots$ (first $4$ terms)
$(2)$ $2,3, \frac{9}{ 2 }, \ldots$ (first $5$ terms)
$(3)$ $100,20,4, \ldots$ (first $5$ terms)
$(4)$ $\frac{1}{ 2 ^{\prime}}, \frac{1}{ 4 ^{\prime}}, \frac{1}{ 8 ^{\prime}}, \ldots$ (first $10$ terms)
Answer$(1)$ $4,16,64, \ldots S_4=$ ?
Here, $a=4 ; r=\frac{16}{4}=4 ; n=4$
$S _{ n }=\frac{ a \left[ r ^{ n }-1\right]}{ r -1}$
$\therefore S _4=\frac{4\left[(4)^4-1\right]}{4-1}=\frac{4[256-1]}{3}=\frac{4 \times 255}{3}=4 \times 85=340$
$(2)$ $2,3, \frac{9}{2^{\prime}} \ldots S_5=$ ?
Here, $a=2, r=\frac{3}{2} ; n=5$
$S _{ n }=\frac{a\left[ r ^{ n }-1\right]}{ r -1}$
$\therefore S _4=\frac{2\left[\left(\frac{2}{2}\right)^5-1\right]}{\frac{3}{2}-1}=\frac{2\left[\frac{243}{32}-1\right]}{\frac{1}{2}}=2 \times 2\left[\frac{243-32}{32}\right]=\frac{211}{8}$
$(3)$ $100,20,4, \ldots S_5=$ ?
Here, $a=100 ; r=\frac{20}{100}=\frac{1}{5^{\prime}} n=5$
$S_n=\frac{a\left[1-r^n\right]}{1-r}$
$\therefore S_5=\frac{100\left[1-\left(\frac{1}{5}\right)^5\right]}{1-\frac{1}{5}}=\frac{100\left[1-\frac{1}{3125}\right]}{\frac{4}{5}}=100 \times \frac{5}{4}\left[\frac{3125-1}{3125}\right]=125 \times \frac{3124}{3125}=\frac{3124}{25}=124.96$
$(4)$ $\frac{1}{2}, \frac{1}{4^{\prime}} \frac{1}{8^{\prime}} \ldots S_{10}=$ ?
Here, $a=\frac{1}{2^{\prime}} r =\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{4} \times \frac{2}{1}=\frac{1}{2^{\prime}} n =10$
$S_{ n }=\frac{ a \left[1- r ^{ n }\right]}{1- r }$
$\therefore S _{10}=\frac{\frac{1}{2}\left[1-\left(\frac{1}{2}\right)^{20}\right]}{1-\frac{1}{2}}=\frac{\frac{1}{2}\left[1-\frac{1}{1024}\right]}{\frac{1}{2}}=\frac{1}{2} \times \frac{2}{1}\left[\frac{1024-1}{1024}\right]=\frac{1023}{1024}$
View full question & answer→Question 243 Marks
If for a $G.P.$ $T_n=324, S_n=484$ and $r=3$; find $a$ and $n$.
Answer$T_n=324, S_n=484, r=3, a=? n=$ ?
Now, $S_n=\frac{r T_n-a}{r-1}$
$\therefore 484=\frac{3(324)-a}{3-1}$
$\therefore 484=\frac{972-a}{2}$
$\therefore 484 \times 2=972-a$
$\therefore a=972-968$
$\therefore a=4$
Now, $T_n=$ a.r ${ }^{n-1}$
$\therefore 324=4(3)^{n-1}$
$\therefore \frac{324}{4}=(3)^{n-1}$
$\therefore 81=(3)^{n-1}$
$\therefore(3)^4=(3)^{n-1}$
$\therefore 4= n -1 \therefore n =5(\because$ Base are equal, index are also equal $)$
View full question & answer→Question 253 Marks
How many terms of a geometric progression $2, 4, 8, 12, ....$ would add to $126 ?$
AnswerHere, $G.P.$ is $2,4,8,16,.........$.
$\therefore a=2, r=\frac{4}{2}=2, S_n=126, n=?$
Now,$S_n=\frac{a\left[r^n-1\right]}{r-1}(\because r > 1)$
$\therefore 126=\frac{2\left[2^n-1\right]}{2-1}$
$\therefore \frac{126}{2}=2^n-1$
$\therefore 63+1=2^n$
$\therefore 64=2^n$
$\therefore(2)^6=2^n$
$\therefore n=6$
View full question & answer→Question 263 Marks
Find the product of the first three terms of a $G.P.$ whose second term is $5.$
Answer$T_2=5 \therefore 5=$ ar
The product of the first three terms of a $G.P.$
$=T_1 \times T_2 \times T_3$
$=a \times a r \times a r^2$
$=a^3 r^3=(a r)^3$
Putting ar $=5$,
The product of the first three terms of a $G.P.$ $(5)^3=125$.
View full question & answer→Question 273 Marks
If $S_n=4\left(3^n-1\right)$, find $T_{n+1}$
Answer$S_n=4\left(3^n-1\right.$
Now, $T _{ n +1}= S _{ n +1}- S _{ n }$
$=4\left[3^{n+1}-1\right]-4\left[3^n-1\right]$
$\left.=4\left[3^{n+1}-1\right]-3^n+1\right]$
$=4\left[3^n(3-1)\right]$
$=4\left(3^n \times 2\right)$
$\therefore T_{n+1}=8\left(3^n\right)$
View full question & answer→Question 283 Marks
If $S_n=\frac{2}{ 3 }\left(2^n-1\right)$, find $T_4$.
Answer$S_n=\frac{2}{3}\left(2^n-1\right)$
Now, $T_{n+1}=S_{n+1}-S_n$
$=\frac{2}{3}\left[2^{n+1}-1\right]-\frac{2}{3}\left[2^n-1\right]$
$=\frac{2}{3}\left[2^{n+1}-1-2^n+1\right]$
$=\frac{2}{3}\left[2^n(2-1)\right]$
$\therefore T_{n+1}=\frac{2}{3} \times 2^n$
We have to find 4 th term. Therefore, put $n =3$
$T _4=\frac{2}{3} \times 2^3$
$=\frac{2}{3} \times 8=\frac{16}{3}$
View full question & answer→Question 293 Marks
For a given $G.P.$, if $T_5=405$ and $T_7=3645$; then find $T_4$.
Answer$T _5=405, T_7=3645 ; T _4=?$
$T_n=a . r^{n-1}$
$\therefore T_5=a . r^4$
$\therefore 405=\text { a.r }{ }^4.........(1)$
$T_7= a . r ^6$
$\therefore 3645=\text { a.r }{ }^6.........(2)$
$T_4= a . r ^3..........(3)$
Taking the ratio of results $(2)$ and $(1),$
$\frac{3645}{405}=\frac{a r^6}{a r^4}$
$\therefore 9=r^2$
$\therefore r= \pm 3$
Putting $r= \pm 3$ in result $(1),$
$405= a ( \pm 3)^4$
$\therefore 405= a \times 81$
$\therefore a =\frac{405}{81}=5$
Putting $a=5$ and $r= \pm 3$ in result $(3),$
$\therefore T_4=5( \pm 27)$
$\therefore T_4= \pm 135$
View full question & answer→Question 303 Marks
The first term of a $G.P.$ is $\frac{-2}{3}$ and the product of its first three terms is $8$ . Find its $7$ th term.
View full question & answer→Question 313 Marks
The $2\ nd$ term of a $G.P.$ is $20$ and its $5\ th$ term is $\frac{4}{25}$. Find the $8\ th$ term of the progression.
Answer$a=100, r=\frac{1}{5}, T_{8}=\frac{4}{3125}$
View full question & answer→Question 323 Marks
The first term of a $G.P.$ is $\frac{1}{5}$ and the product of its first three terms is $\frac{1}{1000}$. Find its $5\ th$ term.
Answer$r=\frac{1}{2}, T_{5}=\frac{1}{80}$
View full question & answer→Question 333 Marks
The first term of a $G.P.$ is $-625$ and its common ratio is $\frac{-1}{5}$. If the $n$th term of that progression is ${ }_{3125}^{1}$ find $n$.
View full question & answer→Question 343 Marks
The third term of a $G.P.$ is $16$ and its $6th$ term is $1024 .$ Find the $11th$ term of the progression.
Answer$a=1, r=4, T_{11}=(4)^{1} 0=1048576$
View full question & answer→Question 353 Marks
The first term of a $G.P.$ is $6$ and its common ratio is $\frac{1}{2}$. If its nth term is $\frac{3}{512}$, find $n$.
View full question & answer→Question 363 Marks
The first term of a $G.P.$ is $10,000$ and its common ratio is $0.1$. If its nth term is $0.001$, find $n.$
View full question & answer→Question 373 Marks
The first term of a $G.P.$ is $54$ and its common ratio is $\frac{1}{3}$. If the nth term of the progression is $\frac{2}{27}$, find $n$. The first term of a $G.P.$ is $3$ and its common ratio is $2$. If its nth term is $768$, find $n$.
View full question & answer→Question 383 Marks
The fifth term of a $G.P.$ is square of its second term. If its seventh term is $256$, find the first term and common ratio of the $G.P.$
View full question & answer→Question 393 Marks
The first term of a $G.P.$ is $1$ . If its $6th$ term is $-3125$, find its common ratio and $T_{4}$.
Answer$r=-5$, $T_{4}=-125$
View full question & answer→Question 403 Marks
The fourth term of a $G.P.$ is $1000$ and its common ratio is $\frac{5}{2}$. Find its first term.
View full question & answer→Question 413 Marks
The second term of a $G.P.$ is $1$ and its fifth term is $27 .$ Find its first term and common ratio.
Answer$a=\frac{1}{3}, r=3$
View full question & answer→Question 423 Marks
The third term of a $G.P.$ is $9$ and its $6th$ term is $\frac{243}{8}$. Find its first term and common ratio.
Answer$a=4, r=\frac{3}{2}$
View full question & answer→Question 433 Marks
In a $G.P.$, sum of first three terms is $2$ . If the first term is $2$ , find the common ratio.
View full question & answer→Question 443 Marks
Three positive numbers $3, k+3$ and $4 k$ are in $G.P.$ Find the value of $k$.
View full question & answer→Question 453 Marks
For the l. $4,2,1, \frac{1}{2}, \ldots$ the sum of how many terms will be $\frac{31}{4} ?$
View full question & answer→Question 463 Marks
If the second term of a $G.P.$ is $2$. Find the product of first three terms of the $G.P.$
View full question & answer→Question 473 Marks
If the sum of first five terms of a $G.P.$ is $1$ and its first term is $1$, find the common ratio of the $G.P.$
View full question & answer→Question 483 Marks
For a $G.P.$ the sum of three consecutive terms is $27$ and their product is $27$, find the teams.
Answer$9, 3, 1$ $OR$ $1, 3, 9$
View full question & answer→Question 493 Marks
A person donates $Rs\ 5000$ in Jan. $2016$ and $Rs\ 7500$ in Feb. $2016$. If from March $2016$ to June, $2016$ he donates the amount $1.25$ times the amount donated previous months. What amount will he donate in June, $2016$?
View full question & answer→Question 503 Marks
For a $G.P.$ first term $= -2$ and common ratio is $- 1$. Find its 10th term and sum of first $10$ terms.
View full question & answer→