Question 11 Mark
The first term and the fourth term of a $GP.$ are $5$ and $40$ respectively; find the common ratio of a $GP$.
AnswerIn this $G.P.$, the first term $a=5$, the fourth term is 40 i.e. $\mathrm{T}_{4}=40$ and we want to find the common ratio of the $G.P.$ i.e. $r$
Here, $\mathrm{T}_{4}=40$
$\therefore a r^{4-1}=40$ $\left(\because \mathrm{T}_{n}=a r^{n-1}\right)$
$\therefore 5 \times \mathrm{r}^{3}=40$ $(\because a=5)$
$\therefore r^{3}=\frac{40}{5}=8$
$\therefore r=2$
Hence, the common ratio of the $G.P.$ is $2$ .
View full question & answer→Question 21 Mark
The common ratio and the fifth term of a $GP.$ are $3$ and $324$ respectively, find the first term of the $GP$.
AnswerIn the given $G.P.$, the common ratio $r=3$, the fifth term is $324$
i.e. $\mathrm{T}_{5}=324$ and we require the first term of the $G.P.$
i.e. $a$. Here, $\mathrm{T}_{5}=324$
$\therefore \quad a r^{5-1}=324$ $~~~~~~~~~\left(\because \mathrm{T}_{n}=a r^{n-1}\right)$
$\therefore \quad a(3)^{4}=324$ $~~~~~~~~~~(\because r=3)$
$\therefore \quad a \times 81=324$
$\therefore \quad a=\frac{324}{81}=4$
Hence, the first term of the $G.P.$ is $4$ .
View full question & answer→Question 31 Mark
If the first term and the common ratio of a $GP.$ are $7$ and $2$ respectively, find its sixth term.
AnswerIn a given $G.P.$, the first term $a=7$, common ratio $r=2$ and we require the sixth term of $G.P.$ i.e. $n=6$.
Putting values of $a, r$ and $n$ in the general term $\mathrm{T}_{n}=a r^{n-1}$, we get,
$ \mathrm{T}_{6} =7 \times(2)^{6-1}$
$=7 \times(2)^{5}$
$=7 \times 32$
$=224 $
Hence, the sixth term of the $G.P.$ is $224 .$
View full question & answer→Question 41 Mark
State whether the statement " $\mathrm{T}_1=\mathrm{S}_1$ " is true or false in $G.P.$
AnswerFor G.P. $T_1=$ First term, $S_1=$ Sum of first term $=T_1$.
Hence, the statement " $T_1=S_1$ " is true.
View full question & answer→Question 51 Mark
State whether the statement "if $a, b, c, d$ are in $G.P$., then $a d=b c$ " is true or false.
Answer$a, b, c, d$ are in $G.P.$
$\therefore \frac{ b }{ a }=\frac{ c }{ b }=\frac{ d }{ c } \therefore \frac{ b }{ a }=\frac{ d }{ c } \therefore ad = bc$ is true.
View full question & answer→Question 61 Mark
For a $G.P$., if $S_7=15$ and $S_6=11$, then find the seventh term of the $G.P.$
Answer$S_6=15, S_5=11$
Now, $T_{n+1}=S_{n+1}-S_n$. We have to find $6^{\text {th }}$ term. Therefore, put $n=5$.
$\therefore T_6=S_6-S_5=15-11=4$
View full question & answer→Question 71 Mark
For a $G.P$., sum of any two consecutive terms is zero, then what will be the common ratio ?
AnswerSum of two consecutive terms $=0$
$\therefore T_1+T_2=0$
$\therefore a + ar =0$
$\therefore a (1+ r )=0$
$\therefore 1+ r =0$
$r =-1$
Hence, Common ration will be $r=-1$.
View full question & answer→Question 81 Mark
The numbers $4, 1, y$ are in $G.P.$ Find the value of $y.$
AnswerNumbers $4, 1, y$ are in $G.P.$
$\therefore \frac{1}{4}=\frac{y}{1} \therefore 4 y=1 \therefore y=\frac{1}{4}$
View full question & answer→Question 91 Mark
If in a G.P, the $n$th term is given as $T_n=2^{n+1}$, find the common ratio.
Answer$T_n=2^{n+1}$ For G.P. $T_n=a \cdot r^{n-1}$. Comparing these two common ratio $=2$
View full question & answer→Question 101 Mark
Find the sum of twenty terms of the $G.P. 7, 7, 7,.....$
Answer$G.P. 7, 7, 7....$
Here, $a=7 ; r=1 ; n=20 . \therefore S_n=n a$
$\therefore S_{20}=20 \times 7=140$
View full question & answer→Question 111 Mark
Find the common ratio of the $G.P. 0.1, 0.01, 0.001,....$
AnswerThe common ratio of the $G.P. 0.1, 0.01, 0.001$ is $r = \frac{0.01}{0.1} = 0.1$
View full question & answer→Question 121 Mark
What is the $n$th term of the $G.P.$ $a r, a r^2, \mathrm{ar}^3, \ldots$ ?
AnswerThe $n$th term of the $G.P.$ $a r, a r^2, a r^3, \ldots$ is a. $r^n$.
View full question & answer→Question 131 Mark
The first term and common ratio $G.P.$ are respectively $1$ and $3$. If its $7th$ term is $729$, find the sum of first seven terms the $G.P.$
View full question & answer→Question 141 Mark
For the $G.P.$ $1,2,4,8, \ldots$, find ratio of sum of $7 th$ and $3 \mathrm{rd}$ term to the difference of $7$ th and $3 rd$ term.
View full question & answer→Question 151 Mark
For a $G.P. 3,3,3, \ldots ;$ find $\left(S_{5}-S_{3}\right):\left(S_{5}+S_{3}\right)$.
View full question & answer→Question 161 Mark
In a $G.P. T_{1}=\frac{8}{27}$ and $T_{2}=-\frac{4}{9}$. Find the common ratio.
View full question & answer→Question 171 Mark
If the sum of first $\mathrm{n}$ terms of a $G.P.$ is $S_{n}=3\left(2^{n}-1\right)$, find the first term.
View full question & answer→Question 181 Mark
First term of a $G.P.$ is $2$ and common ratio is $4$ . Find its $(n+1)$ th term.
View full question & answer→Question 191 Mark
In a G.P. $5 . T_{5}: T_{3}=4: 5$, find $r$.
View full question & answer→Question 201 Mark
If $a=r=-2 .$ Find the general term of $G.P.$
View full question & answer→Question 211 Mark
Three positive numbers $4, G, 25$ are in $G.P.$ Find $G.$
View full question & answer→Question 221 Mark
If $T_{n}=3^{n-1}$, find the value of $T_{1}+T_{2}$.
View full question & answer→Question 231 Mark
If $S_{n}=3\left(2^{n}-1\right)$, find $S_{8}$
View full question & answer→Question 241 Mark
Find the common ratio for the $G.P. -625,-250,-100$,
AnswerHere, $T_{1}=-625$ and $T_{2}=-250$
Hence, $r=\frac{T_{2}}{T_{1}}=\frac{250}{-625}=\frac{2}{5}$
View full question & answer→Question 251 Mark
Write the assumption for the first four terms of a $G.P.$ and state the common ratio for it.
AnswerThe first four terms of a $G.P.$ are assumed as $\frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}$ Here, common ratio $=r^{2}$
View full question & answer→Question 261 Mark
Write the series formula when $|r|>1$, a and $T_{n}$ are given.
AnswerFor $|\mathbf{r}|>1, S_{n}=\frac{r T_{n}-a}{r-1}$
View full question & answer→Question 271 Mark
Write the series formula for $|r|>1$.
AnswerFor $|r|>1, S_{n}=\frac{a\left|r^{n}-1\right|}{r-1}$
View full question & answer→Question 281 Mark
When $|r|<1$, a and $T_{n}$ are given, write the series formula.
AnswerFor $|r|<1, S_{n}=\frac{a-r T_{n}}{1-r}$
View full question & answer→Question 291 Mark
If $a=\frac{1}{4}$ and $r=1$, what will be the sum of its first $20$ terms?
Answer$a=\frac{1}{4}, r=1$ and $n=20$
If $r=1, S_{n}=$ n.a. Hence, $S_{20}=20 \times \frac{1}{4}=5$
View full question & answer→Question 301 Mark
If the first term of a $G.P.$ is $4$ and its common ratio is $2$, find $S_{1}$
AnswerIn the geometric progression $T_{1}=S_{1}$. Hence, $S_{1}=4$.
View full question & answer→Question 311 Mark
Write the series formula if the common ratio is $|r|<1$.
AnswerIf $|r|<1$, then $S_{n}={ }_{n}\left[\begin{array}{c}\left.1 r^{n}\right] \\ 1-r\end{array}\right].$
View full question & answer→Question 321 Mark
How will you obtain the sequence formula from the series formula of a $G.P.$?
AnswerIf series formula $=S_{n}$ and sequence formula $=T_{n}$, then $T_{n+1}=S_{n+1}-S_{n}$ where, $n \geq 1$ and $T_{1}=S_{1}$.
View full question & answer→Question 331 Mark
State the definition of a geometric series.
AnswerIf the sequence of terms of the $G.P.$ is expressed in the form of $'a'$ and $'r'$, then $S_{n}=a+a r+a r^{2}+\ldots+$
$a r^{n-1}$ is called a geometric series.
View full question & answer→Question 341 Mark
What is the sequence formula of a $G.P.$? Write that formula.
AnswerThe formula to find the $\mathrm{n}$ th term of a $G.P.$ is called its sequence formula. The formula is as follows:
$T_{n}=a \cdot r^{n-1}$
where a = first term, $r=$ common ratio
View full question & answer→Question 351 Mark
Write the terms of the geometric progression in regular succession whose first term $=$ a and common ratio $=r$.
AnswerThe terms in regular succession of the progression whose first term $=a$ and common ratio $=r$ are $a, a r$, $a r^{2}, a r^{3}$
View full question & answer→Question 361 Mark
Write the definition of $G.P.$
AnswerIf ' $a$ ' and ' $r$ ' are the non-zero real numbers, then for $n \geq 1$ the sequence whose nth term is $T_{n}=a \cdot r^{n-1}$ is called Geometric progression.
View full question & answer→Question 371 Mark
Which peculiarity does a $G.P.$ possess?
AnswerA $G.P.$ possesses the following peculiarity. "If the first term of a $G.P.$ and a non-zero constant ratio are given, all the terms of the progression can be obtained."
View full question & answer→Question 381 Mark
For $n \geq 1$, the ratio of which terms of a $G.P.$ is a nonzero constant?
AnswerFor $a \geq 1$, the ratio of $(n+1)$ th term and nth term of a $G.P.$ is a non-zero constant.
View full question & answer→