Question 13 Marks
A geometric progression has common ratio $= 3$ and last term $= 486.$ If the sum of its terms is $728;$ find its first term.
AnswerFor a G.P.,
common ratio, $r = 3 (r > 1)$
Last term, $l = 486$
$S = 728$
$\Rightarrow \frac{l r-a}{r-1}=728$
$\Rightarrow \frac{486 \times 3-a}{3-1}=728$
$\Rightarrow \frac{1458-a}{2}=728$
$\Rightarrow 1458 - a = 1456$
Hence, the first term is $2.$
View full question & answer→Question 23 Marks
A boy spends Rs.$10$ on first day, Rs.$20$ on second day, Rs.$40$ on third day and so on. Find how much, in all, will he spend in $12$ days?
AnswerAmount spent on $1^{st}$ day = Rs. 10
Amount spent on $2^{st}$ day = Rs. 20
Amount spent on $3^{st}$ day = Rs. 40 and so on
Now, $\frac{20}{10}=2, \frac{40}{20}=2$,
Thus, 10, 20, 40, ...... is a G.P. with first term, a = 10 and common ratio, r = 2 (r > 1)
∴ Total amount spent in 12 days = $S_{12}$
$S _n=\frac{a\left(r^n-1\right)}{r-1}$
$\Rightarrow S _{12}=\frac{10\left(2^{12}-1\right)}{2-1}=10\left(2^{12}-1\right)=10(4096-1)=10 \times 4095=40950$
Hence, the total amount spent in 12 days is Rs. 40950.
View full question & answer→Question 33 Marks
How many terms of the geometric progression $1 + 4 + 16 + 64 + ……..$ must be added to get sum equal to $5461?$
AnswerGiven G.P. $: 1 + 4 + 16 + 64 +...........$
Here,
first term, $a = 1$
common ratio, $r=\frac{4}{1}=4(r>1)$
Let the number of terms to be added $= n$
Then, $S_n = 5461$
$\Rightarrow \frac{a\left(r^n-1\right)}{r-1}=5461 $
$ \Rightarrow \frac{1\left(4^{\text {th }}-1\right)}{4-1}=5461 $
$ \Rightarrow \frac{4^{\text {th }}-1}{3}=5461$
$ \Rightarrow 4^{\text {th }}-1=16383$
$ \Rightarrow 4^{\text {th }}=16384 $
$\Rightarrow n =7$
Hence, required number of terms $= 7$
View full question & answer→Question 43 Marks
Find the sum of the sequence $-\frac{1}{3}, 1,-3,9, \ldots . . . .$. upto $8$ terms.
AnswerHere,$\frac{1}{-\frac{1}{3}}=\frac{-3}{1}=\frac{9}{-3}=-3$
Thus, the given sequence is a G.P. with First term $(a)=-\frac{1}{3}$ and common ratio $(r)=-3(r<1)$.
Number of terms to be added, $n = 8$
$\therefore S_n=\frac{a\left(1-r^n\right)}{1-r} $
$ \Rightarrow S_s=\frac{-\frac{1}{3}\left(1-(-3)^s\right)}{1+3}=\frac{-1+3^s}{12}=\frac{1}{12}\left(3^s-1\right)$
View full question & answer→Question 53 Marks
The first two terms of a G.P. are $125$ and $25$ respectively. Find the $5^{th}$ and the $6^{th}$ terms of the G.P.
AnswerFirst term (a) = 125
And, common ratio (r) $=\frac{25}{125}=\frac{1}{5}$
Now $t_n=a r^{n-1}$
$\Rightarrow 5^{t h}$ term $=t_5=125 \times\left(\frac{1}{5}\right)^{5-1}=125 \times\left(\frac{1}{5}\right)^4=125 \times \frac{1}{625}=\frac{1}{5}$
$\Rightarrow 6^{\text {th }}$ term $=t_6=125 \times\left(\frac{1}{5}\right)^{6-1}=125 \times\left(\frac{1}{5}\right)^5=125 \times \frac{1}{3125}=\frac{1}{25}$
View full question & answer→Question 63 Marks
Find the sum of G.P.: $\sqrt{3}+\frac{1}{\sqrt{3}}+\frac{1}{3} \sqrt{3}+\ldots . .$. to $n$ terms.
AnswerGiven G.P. : $\sqrt{3}+\frac{1}{\sqrt{3}}+\frac{1}{3 \sqrt{3}}+\ldots \ldots \ldots \ldots . .$. upto $n$ terms
Here,
First term, a $=\sqrt{3}$
common ratio, $r =\frac{\frac{1}{\sqrt{3}}}{\sqrt{3}}=\frac{1}{3}(r<1)$
number of terms to be added $= n$
$\therefore S _n=\frac{a\left(1-r^n\right)}{1-r} $
$ \Rightarrow S _n=\frac{\sqrt{3}\left(1-\left(\frac{1}{3}\right)^n\right)}{1-\frac{1}{3}} $
$ =\frac{\sqrt{3}\left(1-\frac{1}{3^n}\right)}{\frac{2}{3}} $
$ =\frac{3 \sqrt{3}}{2}\left(1-\frac{1}{3^n}\right)$
View full question & answer→Question 73 Marks
Find the 5 th term of the G.P. $\frac{5}{2}, 1$ ..........
AnswerFirst term $( a )=\frac{5}{2}$
And, common ratio $(r)=\frac{1}{\frac{5}{2}}=\frac{2}{5}$
Now $t_n=a r^{n-1}$
$\Rightarrow 5^{t h}$ term $=t_5=\frac{5}{2} \times\left(\frac{5}{2}\right)^{5-1}=\frac{5}{2} \times\left(\frac{2}{5}\right)^4=\left(\frac{2}{3}\right)^3=\frac{8}{125}$
View full question & answer→Question 83 Marks
Find the sum of G.P.: $1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\ldots \ldots . .$. to $n$ terms
AnswerGiven G.P. : $1+\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\ldots$. upto $n$ terms
Here,
first term, $a = 1$
common ratio, $r=\frac{-\frac{1}{3}}{1}=-\frac{1}{3}(r<1)$
number od terms to be added $= n$
$\therefore S _n=\frac{a\left(1-r^n\right)}{1-r} $
$\Rightarrow S _n=\frac{1\left(1-\left(-\frac{1}{3}\right)^n\right)}{1-\left(-\frac{1}{3}\right)} $
$ =\frac{1\left(1-\left(-\frac{1}{3}\right)^n\right)}{1+\frac{1}{3}} $
$ =\frac{1-\left(-\frac{1}{3}\right)^n}{\frac{4}{3}} $
$ =\frac{3}{4}\left[1-\left(-\frac{1}{3}\right)^n\right]$
View full question & answer→Question 93 Marks
Find the sum of G.P.:
0.3 + 0.03 + 0.003 + 0.0003 +….. to 8 items.
AnswerGiven G.P. : 0.3 + 0.03 + 0.003 + 0.003 +...........
Here,
first term, a = 0.3
common ratio, $r=\frac{0.03}{0.3}=0.1(r<1)$
number of ter to be added, n = 8
$\therefore S _n=\frac{a\left(1=r^n\right)}{1-r}$
$\Rightarrow S_8 \frac{0.3\left(1-(0.1)^8\right)}{1-0.1}=\frac{0.3\left(1-(0.1)^8\right)}{0.9}=\frac{1-(0.1)^8}{3}=\frac{1}{3}\left(1-\frac{1}{10^8}\right)$
View full question & answer→Question 103 Marks
Find the sum of G.P.:
1 + 3 + 9 + 27 + ………. to 12 terms
AnswerGiven G.P. : 1 + 3 + 9 + 27+.......
Here,
first term, a = 1
common ratio, $r=\frac{3}{1}=3(r>1)$
number of terms to be added, n = 12
$\therefore S_n=\frac{a\left(r^n-1\right)}{r-1}$
$S _{12}=\frac{1\left(3^{12}-1\right)}{3-1}=\frac{3^{12}-1}{2}=\frac{531441-1}{2}=\frac{531440}{2}=265720$
View full question & answer→Question 113 Marks
If a, b and c are in A.P, a, x, b are in G.P. whereas b, y and c are also in G.P.
Show that: $x^2, b^2, y^2$ are in A.P.
Answera, b and c are in AP.
$\Rightarrow 2b = a + c$
a, x and b are in GP.
$\Rightarrow x^2 = ab$
b, y and c are in GP.
$\Rightarrow y^2 = bc$
now.
$x^2 + y^2 = ab + bc$
$= b(a+ c)$
$= b x 2b$
$= 2b^2$
$\Rightarrow x^2, b^2$ and $y^2$ are in A.P.
View full question & answer→Question 123 Marks
If each term of a G.P. is raised to the power x, show that the resulting sequence is also a G.P.
AnswerLet $a_1, a_2, a_3, ................., a_n, ..........$be a G.P. with common ratio r.
$\Rightarrow \frac{a_{n+1}}{a_n}=r$ for all $n \in N$
If each term of a G.P. is raised to the power $x$, we get the sequence $a _1^x, a _2^x, a _3^x, \ldots . . . . ., a _n^x, \ldots . . .$.
Now, $\frac{\left(a_{n+1}\right)^x}{\left(a_n\right)^x}=\left(\frac{a_{n+1}}{a_n}\right)^x=r^x$ for all $n \in N$
Hence, $a _1^x, a _2^x, a _3^x, \ldots . . . . ., a _n^x, \ldots . . .$. is also a G.P.
View full question & answer→Question 133 Marks
If for a G.P., $p^{th}, q^{th}$ and $r^{th}$ terms are a, b and c respectively ; prove that: $(q - r) log a + (r - p) log b + (p + q) log c = 0$
AnswerLet the first term of the G.P. be A and its common ratio be R.
Then,
$p^{th}$ term $= a \Rightarrow AR^{p-1} = aq^{th}$ term $= b \Rightarrow AR^{q-1} = b$
$r^{th}$ term $= c \Rightarrow AR^{r-1} = c$
Now,
$a^{q-r} x b^{r-p} x c^{p-q} = (AR^{p-1})^{q-r} x (AR^{q-1})^{r-p} x (AR^{r-1})^{p-q}$
$= A^{q-r}. R^{(p-1)(q-r)} x A^{r-p} . R^{(q-1)(r-p)} x A^{p-q} . R^{(r-1)(p-q)}$
$= A^{q-r+r-p+p-q}x R^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}$
$= A^0 x R^0$
$= 1$
Taking log on both the sides, we get
$\log (a^{q-r} x b^{r-p} x c^{p-q}) = log1$
$\Rightarrow (q - r)\log a + (r - p)\log b + (p - q)\log c = 0 ...$(proved)
View full question & answer→Question 143 Marks
For the G.P. $\frac{1}{27}, \frac{1}{9}, \frac{1}{3}, \ldots \ldots 81$
find the product of fourth term from the beginning and the fourth term from the end.
AnswerGiven G.P. : $\frac{1}{27}, \frac{1}{9}, \frac{1}{3}, \ldots \ldots 81$
Here,
Common ratio, r $=\frac{\frac{1}{9}}{\frac{1}{27}}=3$
First term, $a =\frac{1}{27}$ and last term, $I =81$
$\therefore 4^{\text {th }}$ term from the beginning $=a^3=\frac{1}{27} \times(3)^3=\frac{1}{27} \times 27=1$
And, $4^{\text {th }}$ term from an end $=\frac{1}{ r ^3}=\frac{81}{(3)^3}=\frac{81}{27}=3$
Thus, required product = 1 x 3 = 3
View full question & answer→Question 153 Marks
If the $4^{th}$ and $9^{th}$ terms of a G.P. are $54$ and $13122$ respectively, find the G.P. Also, find its general term.
AnswerLet the first term of the G.P. be a and its common ratio be r.
$4^{th} term = t_4 = 54 \Rightarrow ar^3 = 54$
$9^{th} term = t_9 = 13122 \Rightarrow ar^8 = 13122$
$\frac{ ar ^8}{ ar ^3}=\frac{13122}{54}$
$\Rightarrow r^5 = 243$
$\Rightarrow r = 3$
$ar^3 = 54$
$\Rightarrow a x (3)^3 = 54$
$\Rightarrow a =\frac{54}{27}=2$
$\therefore$ Required $G.P = a, ar, ar^2, ar^3, ........$
$= 2, 2 \times 3, 2 \times (3)^2, 54$
$= 2, 6, 18, 54$
General term $= t_n = ar^{n-1} = 2 x (3)^{n-1}$
View full question & answer→Question 163 Marks
Second term of a Geometric Progression is $6$ and its fifth term is $9$ times of its third term. Find the geometric progression. Consider that each term of the G.P. is positive.
AnswerLet the first term of the G.P. be a and its common ratio be r.
Now, $2^{nd}$ term $= t_2 = 6 \Rightarrow ar = 6$
Also, $t_5 = 9 x t_3$
$\Rightarrow ar^4 = 9 x ar^2$
$\Rightarrow r^2 = 9$
$\Rightarrow r =±3$
Since, each term of a G.P. is positive, we have $r = 3$ and $ar = 6\Rightarrow a x 3 = 6 \Rightarrow a = 2$
$\therefore G.P. = a, ar, ar^2, ar^3, ........$
$= 2, 6, 2 x (3)^2, 2 x (3)^3,............$
$= 2, 6, 18, 54,..........$
View full question & answer→Question 173 Marks
Find the Geometric progression with $4^{th}$ term = $54$ and $7^{th}$ term = $1458$.
AnswerLet the first term of the G.P. be a and its common ratio be r.
$4^{th}$ term = $54 \Rightarrow ar^3 = 54$
$7^{th}$ term = $1458 \Rightarrow ar^6 = 1458$
Now, $\frac{ ar ^6}{ ar ^3}=\frac{1458}{54}$
$\Rightarrow r^3 = 27$
$\Rightarrow r = 3$
$ar^3= 54$
$\Rightarrow a x (3)^3 = 54$
$\Rightarrow a =\frac{54}{27}=2$
$\therefore G.P. = a, ar, ar^2, ar^3,$ ........
$= 2, 2 \times 3, 2 \times (3)^2, 54,$...........
$= 2, 6, 18, 54$,............
View full question & answer→Question 183 Marks
Which term of the G.P. :
$-10, \frac{5}{\sqrt{3}},-\frac{5}{6}, \ldots \ldots \ldots .$. is $-\frac{5}{72} ?$
AnswerFor the given G.P. :
First term, $a =-10$
Common ratio, $r=\frac{\frac{5}{\sqrt{3}}}{-10}=-\frac{1}{2 \sqrt{3}}$
If $-\frac{5}{72}$ is the $n ^{\text {th }}$ term of the geiven G.P., then $-\frac{5}{72}=\operatorname{ar}^{ n -1}$
$\Rightarrow-\frac{5}{72}=-10 \times\left(\frac{1}{2 \sqrt{3}}\right)^{n-1} $
$ \Rightarrow \frac{1}{144}=\left(\frac{1}{2} \sqrt{3}\right)^{n-1}$
$\Rightarrow \frac{1}{2 \times 2 \times 2 \times 2 \times \sqrt{3} \times \sqrt{3} \times \sqrt{3} \times \sqrt{3}}=\left(\frac{1}{2 \sqrt{3}}\right)^{n-1}$
$\Rightarrow\left(\frac{1}{2 \sqrt{3}}\right)^4=\left(\frac{1}{2 \sqrt{3}}\right)^{n-1}$
$\Rightarrow n-1=4 \\ \Rightarrow n=5$
View full question & answer→Question 193 Marks
The fifth, eight and eleventh terms of a geometric progression are p, q and r respectively. show that : $q^2 = pr$.
AnswerLet the first term of the G.P. be a and its common ratio be r .
$5^{\text {th }} \text { term }=\mathrm{t}_5=\mathrm{p}$
$\Rightarrow \mathrm{ar}^4=\mathrm{p}$
$8^{\text {th }} \text { term }=\mathrm{t}_8=\mathrm{q}$
$\Rightarrow \mathrm{ar}^7=\mathrm{q}$
$11^{\text {th }} \text { term }=\mathrm{t}_{11}=\mathrm{r}$
$\Rightarrow \mathrm{ar}^{10}=\mathrm{r}$
Now,
$\mathrm{pr}=\mathrm{ar}^4 \times \mathrm{ar}^{10}=\mathrm{a}^2 \times \mathrm{r}^{14}=\left(\mathrm{a} \times \mathrm{r}^7\right)^2=\mathrm{q}^2$
$\Rightarrow \mathrm{q}^2=\mathrm{pr}$
View full question & answer→Question 203 Marks
Find the seventh term of the G.P. :
$\sqrt{3}+1,1, \frac{\sqrt{3}-1}{2}..............$
AnswerGiven G.P. : $\sqrt{3}+1,1, \frac{\sqrt{3}-1}{2}, \ldots \ldots \ldots$
Here,
First term, $a =\sqrt{3}+1$
Common ratio, $r =\frac{1}{\sqrt{3}+1}$
Now, $t_n = ar^{n-1}$
$\Rightarrow t_7=(\sqrt{3}+1) \times\left(\frac{1}{\sqrt{3}+1}\right)^6 $
$ =\left(\frac{1}{\sqrt{3}+1}\right)^5$
$ =\left(\frac{1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}\right)^5 $
$ =\left(\frac{\sqrt{3}-1}{2}\right)^5$
$=\frac{1}{32}(\sqrt{3}-1)^5$
View full question & answer→Question 213 Marks
Find the next three tearms of the sequence :
$\sqrt{5}, 5,5 \sqrt{5}............$
AnswerGiven sequence : $\sqrt{5}, 5,5 \sqrt{5}$..
Now,
$\frac{5}{\sqrt{5}}=\sqrt{5}, \frac{5 \sqrt{5}}{5}=\sqrt{5}$
Since $\frac{5}{\sqrt{5}}=\frac{5 \sqrt{5}}{52}=\ldots \ldots \ldots . .=\sqrt{5}$,
the given sequence is a G.P. with first term, $a=\sqrt{5}$ and
Now, $t_n = ar^{n-1}$
$\therefore$ Next three term:
$4^{\text {th }} \text { term }=\sqrt{5} \times(\sqrt{5})^3=\sqrt{5} \times 5 \sqrt{5}=25$
$5^{\text {th }} \text { term }=\sqrt{5} \times(\sqrt{5})^4=\sqrt{5} \times 25=25 \sqrt{5}$
$6^{\text {th }} \text { term }=\sqrt{5} \times(\sqrt{5})^5=\sqrt{5} \times 25 \sqrt{5}=125$
View full question & answer→Question 223 Marks
Find the $8^{th}$ term of the sequence:
$\frac{3}{4}, 1 \frac{1}{2}, 3, \ldots \ldots \ldots \ldots$
AnswerGiven sequence: $\frac{3}{4}, 1 \frac{1}{2}, 3, \ldots \ldots \ldots \ldots$
i.e. $\frac{3}{4}, \frac{3}{2}, 3, \ldots \ldots$
Now,
$\frac{\frac{3}{2}}{\frac{3}{4}}=2, \frac{3}{\frac{3}{2}}=2$,
Since $\frac{\frac{3}{2}}{\frac{3}{4}}=\frac{3}{\frac{3}{2}}=\ldots \ldots .=2$ the given sequence is a G.P. with first term,A=3/4 and common ratio, r=2
Now, $t_n = ar^{n-1}$
$\Rightarrow t _8 \frac{3}{4} \times 2^7=\frac{3}{4} \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2=3 \times 2^5=96$
View full question & answer→Question 233 Marks
Find the next three terms of the series:
$\frac{2}{27}, \frac{2}{9}, \frac{2}{3}..............$,
AnswerGiven sequence: $\frac{2}{27}, \frac{2}{9}, \frac{2}{3}$,...............
Now,
$\frac{\frac{2}{9}}{\frac{2}{27}}=3, \frac{\frac{2}{3}}{\frac{2}{9}}=3$
Since $\frac{\frac{2}{9}}{\frac{2}{27}}=\frac{\frac{2}{3}}{\frac{2}{9}}=\ldots . . . . .=3$,
the given sequence is a G.P. with first term, $a=\frac{2}{27}$ and common ratio, $r=3.$
Now, $t_n = ar^{n-1}$
$\therefore$ Next three terms:
$4^{\text {th }} \text { term }=\frac{2}{27} \times(3)^3=\frac{2}{27} \times 27=2$
$\text { th } \text { term }=\frac{2}{27} \times(3)^4=\frac{2}{27} \times 27 \times 3=6$
$t^{\text {th }} \text { term }=\frac{2}{27} \times(3)^5=\frac{2}{27} \times 27 \times 9=18$
View full question & answer→