MCQ 1011 Mark
If in an $A.P.,$ first term is $20,$ common difference is $2$ and nth term is $42$, then find $n.$
AnswerWe know, $a = 20, d = 2, a_n = 42$
$a + (n - 1) d = 42 $
$\Rightarrow 20 + 2(n - 1) = 42$
$\Rightarrow 2 (n - 1) = 42 - 20 = 22 $
$\Rightarrow n - 1 = 11 $
$\Rightarrow n = 12.$
View full question & answer→MCQ 1021 Mark
Which of the following is not a series?
AnswerThe isometric series is not a series.
Rest all are series i.e. arithmetic series, geometric series and harmonic series.
View full question & answer→MCQ 1031 Mark
A person is to count $4500$ currency notes. Let an denotes the number of notes he counts in the nth minute. If $a_1 = a_2 = . . . .= a_{10}= 150$ and $a_{10} , b_{11}$, . . . are in $AP$ with common difference $2,$ then the time taken by him to count all notes, is:
- A
$24$ min
- ✓
$34$ min
- C
$125$ min
- D
$135$ min
AnswerCorrect option: B. $34$ min
View full question & answer→MCQ 1041 Mark
The sum of the series $\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+\frac{80}{81}+\ ....$ to $n$ terms is:
- A
$\text{n}-\frac{1}{2}(3^{-\text{n}}-1)$
- ✓
$\text{n}-\frac{1}{3}(1-3^{-\text{n}})$
- C
$\text{n}+\frac{1}{2}(3^\text{n}-1)$
- D
$\text{n}-\frac{1}{1}(3^\text{n}-1)$
AnswerCorrect option: B. $\text{n}-\frac{1}{3}(1-3^{-\text{n}})$
Let $T_n$ be the $n^{th}$ term of the given series.
Thus, we have
$\text{T}_\text{n}=\frac{3^\text{n}-1}{3^\text{n}}=1-\frac{1}{3^\text{n}}$
Now,
Let $S_n$ be the sum of $n$ terms of the given series.
Thus, we have
$\text{S}_\text{n}=\sum^\limits\text{n}_{\text{k}=1}\text{T}_\text{k}$
$=\sum^\limits\text{n}_{\text{k}=1}\Big[1-\frac{1}{3^\text{k}}\Big]$
$=\sum^\limits\text{n}_{\text{k}=1}1-\sum^\limits\text{n}_{\text{k}=1}\frac{1}{3^\text{k}}$
$=\text{n}-\Big[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ ....\ +\frac{1}{3^\text{n}}\Big]$
$=\text{n}-\frac{1}{3}\Bigg[\frac{1-\big(\frac{1}{3}\big)^\text{n}}{1-\frac{1}{3}}\Bigg]$
$=\text{n}-\frac{1}{3}\Big[1-\Big(\frac{1}{3}\Big)^\text{n}\Big]$
$=\text{n}-\frac{1}{3}\big[1-3^{-\text{n}}\big]$
View full question & answer→MCQ 1051 Mark
Find the sum of first $n$ terms.
- ✓
$\frac{\text{n}(\text{n}+1)}{2}$
- B
$\Big(\frac{\text{n}(\text{n}+1)}{2}\Big)$
- C
$\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
- D
$\Big(\frac{\text{n}(\text{n}+1)}{2}\Big)$
AnswerCorrect option: A. $\frac{\text{n}(\text{n}+1)}{2}$
Sum of first $n$ terms $= 1+2+3+4+……+n$
$\Rightarrow\Big(\frac{\text{n}}{2}\Big)=\text{(a}+\text{b)}$
$=\Big(\frac{\text{n}}{2}\Big)(\text{1}+\text{n})$
$=\frac{\text{n}(\text{n}+1)}{2}.$
View full question & answer→MCQ 1061 Mark
If $a, b, c$ are in $AP,$ then $10^{ax+10}, 10^{bx+10}, 10^{cx}+10(\mathrm{x} \neq 0)$ are in:
- A
$AP$
- B
$GP$ only when $x > 0$
- ✓
$GP$ for all $x$
- D
$GP$ only when $x < 0$
AnswerCorrect option: C. $GP$ for all $x$
View full question & answer→MCQ 1071 Mark
If $a, b, c, d$ are any four consecutive coefficients of any expanded binomial, then $\frac{\text{a}+\text{b}}{a},\frac{\text{b}+\text{c}}{b},\frac{\text{c }+\text{d}}{c}$ are in:
View full question & answer→MCQ 1081 Mark
The sum of $n$ terms of two arithmetic progressions are in the ratio $(2n + 3) : (7n + 5).$ Find the ratio of their $9^{th}$ terms.
- A
$4 : 5$
- B
$5 : 4$
- ✓
$9 : 31$
- D
$31 : 9$
AnswerCorrect option: C. $9 : 31$
Let $\text{a, a’}$ be the first terms and $\text{d, d’}$ be the common differences of $\text{2 A.P.’s}$ respectively.
Given, $\frac{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]}{\frac{\text{n}}2[2a'+(\text{n}-1){\text{d}]}}=\frac{2\text{n}+3}{7\text{n}+5}$
$\Rightarrow\frac{\text{a}+(\text{n}-1){\frac{d}2{}}}{\text{a'}+(\text{n}-\frac{\text{d'}}{2}}{}=\frac{2\text{n}+3}{7\text{n}+5}$
If we have to find ratio of $9^{th}$ terms then
$\frac{(\text{n}-1)}{2}=8$
$\Rightarrow\text{n}=17$
$\Rightarrow\frac{\text{a}+8\text{d}}{\text{a'}+8\text{d'}}$
$=\frac{2\times17+3}{3\times17+5}$
$=\frac{34+3}{119+5}$
$=\frac{36}{124}$
$=\frac{9}{31}.$
View full question & answer→MCQ 1091 Mark
In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then, the common ratio of this progression is equal to:
AnswerCorrect option: C. $\sqrt{5}$
View full question & answer→MCQ 1101 Mark
If second term of a $\text{G.P.}$ is $2$ and the sum of its infinite terms is $8,$ then its first terms is:
- A
$\frac{1}{4}$
- B
$\frac{1}{2}$
- C
$2$
- ✓
$4.$
Answer$\text{a}_2=2$
$\therefore\text{ar}=2\ \cdots(\text{i})$
Also, $\text{S}_\infty=8$
$\Rightarrow\frac{\text{a}}{(1-\text{r})}=8$
$\Rightarrow\frac{\text{a}}{\Big(1-\frac{2}{\text{a}}\Big)}=8 [$Using $(i)]$
$\Rightarrow\text{a}^2=8(\text{a}-2)$
$\Rightarrow\text{a}^2-8\text{a}+16=0$
$\Rightarrow(\text{a}-4)^2=0$
$\Rightarrow\text{a}=4$
View full question & answer→MCQ 1111 Mark
Find the sum of first $5$ terms of series $2 + 4 + 6 +...........$
AnswerSince $2, 4$ and $6$ all are even numbers
so, given series involve all even number terms.
The next two terms will be $8$ and $10$
so, sum will be $2 + 4 + 6 + 8 + 10 = 30.$
View full question & answer→MCQ 1121 Mark
Choose the correct answer. If the third term of $\text{G.P.}$ is $4,$ then the product of its first $5$ terms is:
AnswerGiven that:
$T_3 = 4$
$\Rightarrow ar^{3-1}= 4$
$\big[\because\ \text{T}_\text{n}=\text{ar}^{\text{n}-1}\big]$
$\Rightarrow ar^2 = 4$
Product of first $5$ terms $= \ce{a. ar. ar^2. ar^3. ar^4}$
$= a^5r^{10}= (ar^2)^5= (4)^5$
Hence, the corrrect option is $(c).$
View full question & answer→MCQ 1131 Mark
The AM,HMandGMbetween two numbers are $\frac{144}{15},15$ and 12, but not necessarily in thisorder. Then, HM, GM and AM respectively are:
AnswerCorrect option: B. $ \frac{144}{15},12, 15$
$ \frac{144}{15},12, 15$
View full question & answer→MCQ 1141 Mark
If $100$ times the $100^{th}$ term of an $AP$ with non$-$zero common difference equals the $50$ times its $50^{th}$ term, then the $150^{th}$ term of this $AP$ is:
View full question & answer→MCQ 1151 Mark
The sum of first three terms of a $G.P.$ is $\frac{21}{2}$ and their product is $27.$ Which of the following is not a term of the $G.P.$ if the numbers are positive?
- A
$3$
- ✓
$\frac{2}{3}$
- C
$\frac{3}{2}$
- D
$6$
AnswerCorrect option: B. $\frac{2}{3}$
Let three terms be $\frac{\text{a}}{\text{r}}, a, \times ar.$
$\text{product}=27\Rightarrow\Big(\frac{\text{a}}{\text{r}}\Big)\Big(\text{a}\Big)\Big(\text{a}\times\text{r}\Big)=27$
$\Rightarrow\text{a}^3=27$
$\Rightarrow\text{a}=3.$
$\text{sum}=\frac{21}{2}$
$\Rightarrow\Big(\frac{a}{r+\text{a}+\text{a}\times\text{r}}\Big)=\frac{21}{2}$
$\Rightarrow\text{a}\Big( \frac{1 }{\text{r+1+1}\times\text{r}}\Big) =\frac{21}{2} $
$\Rightarrow\Big(\frac{1}{\text{r+1+1}\times\text{r}}\Big)=\Big(\frac{\frac{21}{2}}{3}\Big)=\frac{7}{2}$
$\Rightarrow\Big(\text{r}^2+\text{r}+1=\Big)\Big(\frac{7}{2}\Big)$
$\Rightarrow\text{r}^2 -\Big(\frac{5}{2}\Big)\text{r}+1=0$
$\Rightarrow\text{r}=2$ and $\frac{1}{2}.$
Terms are $\frac{3}{2}, 3, 3 \times 2 $
i.e. $\frac{3}{2} , 3, 6.$
View full question & answer→MCQ 1161 Mark
The sum of $n$ terms of an $AP$ is a $n(n – 1).$ The sum of the squares of these terms is:
- A
$\text{n}^2-\text{n}^2(\text{n-1)}^2$
- B
$\frac{\text{a}^2}{6}-\text{n}(\text{n-1})(2\text{n-1)}$
- ✓
$\frac{2\text{a}^2}{3}-\text{n}(\text{n-1})(2\text{n-1)}$
- D
$\frac{2\text{a}^2}{3}-\text{n}(\text{n+1})(2\text{n+1)}$
AnswerCorrect option: C. $\frac{2\text{a}^2}{3}-\text{n}(\text{n-1})(2\text{n-1)}$
View full question & answer→MCQ 1171 Mark
Complete $2, 4, 6, 8, .............$
AnswerSince sequence $2, 4, 6, 8, 10$ contains limited number of terms
so, it is finite sequence.
Rest all are infinite sequences.
View full question & answer→MCQ 1181 Mark
If the non$-$zero numbers $\text{x, y, z}$ are in $\text{AP}$ and $\ce{\tan - 1 , \tan - 1 , \tan - 1 x y z}$ are in $\text{AP,}$ then:
- ✓
$x = y = z$
- B
$yzx 2 =$
- C
$xyz 2 =$
- D
$zxy2 =$
AnswerCorrect option: A. $x = y = z$
View full question & answer→MCQ 1191 Mark
In a $\text{G.P.}$ of ever number of terms, the sum of all terms is five times the sum of the odd terms. The common ratio of the $\text{G.P.}$ is:
- A
$-\frac{4}{5}$
- B
$\frac{1}{5}$
- ✓
$4$
- D
AnswerLet there be $2n$ terms in a $G.P.$
Let a be the first term and $r$ be the common ratio.
$\because\text{ S}_{2\text{n}}=5(\text{S}_{\text{odd terms}})$
$\Rightarrow\frac{\text{a}\big(\text{r}^{2\text{n}-1}\big)}{(\text{r}-1)}=5\big(\text{a}+\text{ar}^2+\text{ar}^4+\text{ar}^6+\dots\text{ar}^{(2\text{n}-1)}\big)$
$\Rightarrow\frac{\text{a}\big(\text{r}^{2\text{n}}-1\big)}{(\text{r}-1)}=5\Bigg(\frac{\text{a}\big(\big(\text{r}^2\big)^\text{n}\big)}{\big(\text{r}^2-1\big)}\Bigg)$
$\Rightarrow\frac{\big(\text{r}^{2\text{n}}-1\big)}{(\text{r}-1)}=5\frac{\big(\big(\text{r}^2\big)^\text{n}-1\big)}{\big(\text{r}^2-1\big)}$
$\Rightarrow\frac{\big(\big(\text{r}^\text{n}\big)^2-1^2\big)}{(\text{r}-1)}=5\frac{\big(\big(\text{r}^\text{n}\big)^2-1^2\big)}{\big(\text{r}^2-1\big)}$
$\Rightarrow\frac{\big(\text{r}^\text{n}-1\big)\big(\text{r}^\text{n}+1\big)}{(\text{r}-1)}=5\frac{\big(\text{r}^\text{n}-1\big)\big(\text{r}^\text{n}+1\big)}{\big(\text{r}-1\big)\big(\text{r}+1\big)}$
$\Rightarrow\big(\text{r}^\text{n}-1\big)\big(\text{r}^\text{n}+1\big)\big(\text{r}-1\big)\big(\text{r}+1\big) -5\big(\text{r}-1\big)\big(\text{r}^\text{n}-1\big)\big(\text{r}^\text{n}+1\big)=0$
$\Rightarrow\big(\text{r}^\text{n}-1\big)\big(\text{r}^\text{n}+1\big)\big(\text{r}-1\big)\big(\text{r}+1-5\big)=0$
But, $r = 1$ or $−1$ is not possible.
$\therefore\text{r}=4$
View full question & answer→MCQ 1201 Mark
If sum of $n$ terms of an $A.P.$ is $n^2+5n$ then find general term.
- A
$n + 1$
- ✓
$2n$
- C
$3n$
- D
$n^2+ 3n$
AnswerGiven, $S_n=n^2+5 n$
We know, $a_n=S_n-S_{n-1}$
$=\left(n^2+5 n\right)-\left((n-1)^2+5(n-1)\right)$
$=\left(n^2+5 n\right)-\left(n^2+1-2 n+5 n-1\right)$
$=2 n$.
View full question & answer→MCQ 1211 Mark
The product $(32),(32)^{\frac{1}{6}}(32)^{\frac{1}{36}}\ \dots\text{ to }\infty$ is equal to:
Answer$32\times32^{\frac{1}{6}}\times32^{\frac{1}{36}}\times\ \cdots\infty$
$=32^{\big(1+\frac{1}{6}+\frac{1}{36}+\ \cdots\infty\big)}$
$=32^{\Bigg(\frac{1}{1-\frac{1}{6}}\Bigg)}$
$[\because$ it is a $G.P. ]$
$=32^{\big(\frac65\big)}$
$=\big(2^5\big)^{\big(\frac65\big)}$
$=2^6$
$=64$
View full question & answer→MCQ 1221 Mark
In a $\text{G.P.}$ if the $(m + n)^{th}$ terms is $p$ and $(m - n)^{th}$ term is $q,$ then its $m^{th}$ term is:
AnswerCorrect option: C. $\sqrt{\text{pq}}$
Here, $\text{a}_{(\text{m}+\text{n})}=\text{p}$
$\Rightarrow\text{ar}^{(\text{m}+\text{n}-1)}=\text{p}\ \cdots(\text{i})$
Also, $\text{a}_{(\text{m}-\text{n})}=\text{q}$
$\Rightarrow\text{ar}^{(\text{m}-\text{n}-1)}=\text{q}\ \cdots(\text{ii})$
Mutliplying $(i)$ and $(ii):$
$\Rightarrow\text{ar}^{(\text{m}+\text{n}-1)}\text{ar}^{(\text{m}-\text{n}-1)}=\text{pq}$
$\Rightarrow\text{a}^2\text{r}^{(2\text{m}-2)}=\text{pq}$
$\Rightarrow\Big(\text{ar}^{(\text{m}-1)}\Big)^2=\text{pq}$
$\Rightarrow\text{ar}^{(\text{m}-1)}=\sqrt{\text{pq}}$
$\Rightarrow\text{a}_\text{m}=\sqrt{\text{pq}}$
Thus, the $m^{th}$ term is $\sqrt{\text{pq}}.$
View full question & answer→MCQ 1231 Mark
The sum of first $20$ terms of the sequence $0.7, 0.7 7, 0.7 7 7, … ,$ is:
- A
$7/81 (179 – 10^{-20})$
- B
$7/9 (99 – 10^{-20})$
- ✓
$7/81 (179 + 10^{-20})$
- D
$7/9 (99 + 10^{-20})$
AnswerCorrect option: C. $7/81 (179 + 10^{-20})$
View full question & answer→MCQ 1241 Mark
If $1+\frac{1+2}{2}+\frac{1+2+3}{3}+\ ....$ to $n$ terms is $S,$ then $S$ is equal to:
- ✓
$\frac{\text{n}(\text{n}+3)}{4}$
- B
$\frac{\text{n}(\text{n}+2)}{4}$
- C
$\frac{\text{n}(\text{n}+1)(\text{n}+2)}{6}$
- D
$\text{n}^2$
AnswerCorrect option: A. $\frac{\text{n}(\text{n}+3)}{4}$
Let $T_n$ be the nth term of the given series.
Thus, we have
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\Big(\frac{\text{k}}{2}+\frac{1}{2}\Big)$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\frac{\text{k}}{2}+\frac{\text{n}}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{4}+\frac{\text{n}}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\Big(\frac{\text{n}+1}{2}+1\Big)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{2}\Big(\frac{\text{n}+3}{2}\Big)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+3)}{4}$
View full question & answer→MCQ 1251 Mark
The sum of the series $1^2 + 3^2 + 5^2 + ...$ to $n$ terms is:
- A
$\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{2}$
- ✓
$\frac{\text{n}(2\text{n}-1)(2\text{n}+1)}{3}$
- C
$\frac{(\text{n}-1)^2(2\text{n}+1)}{6}$
- D
$\frac{(2\text{n}+1)^3}{3}$
AnswerCorrect option: B. $\frac{\text{n}(2\text{n}-1)(2\text{n}+1)}{3}$
Let $T_n$ be the $n^{th}$ term of the given series.
Thus, we have
$\text{T}_\text{n}=(2\text{n}-1)^2$
$=4\text{n}^2+1-4\text{n}$
Now, let $S_n$ be the sum of $n$ terms of the given series.
Thus, we have
$\text{S}_\text{n}=\sum\limits^{\text{n}}_\text{k=1}(4\text{k}^2+1-4\text{k})$
$\Rightarrow\text{S}_\text{n}=4\sum\limits^{\text{n}}_\text{k=1}\text{k}^2+\sum\limits^{\text{n}}_\text{k=1}1-4\sum\limits^{\text{n}}_\text{k=1}\text{k}$
$\Rightarrow\text{S}_\text{n}=\frac{4\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\text{n}-\frac{4\text{n}(\text{n}+1)}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{2\text{n}(\text{n}+1)(2\text{n}+1)}{3}+\text{n}-2\text{n}(\text{n}+1)$
$\Rightarrow\text{S}_\text{n}=\text{n}\Big[\frac{2(\text{n}+1)(2\text{n}+1)}{3}+1-2(\text{n}+1)\Big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{3}\big[(2\text{n}+2)(2\text{n}+1)+3-6(\text{n}+1)\big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{3}\big[(4\text{n}^2-1)\big]$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(2\text{n}-1)(2\text{n}+1)}{3}$
View full question & answer→MCQ 1261 Mark
If in an $A.P.,$ first term is $20$ and $12^{th}$ term is $120.$ Find the sum up to $12^{th}$ term.
AnswerGiven, $a = 20, a_{12}= 120, n = 12$
$\text{s}_{\text{n}}=\frac{\text{n}}{2}(\text{a}+\text{I})$
$\Rightarrow\text{s}_{\text{12}}=\frac{12}{2}(20+120)$
$= 6 \times 140$
$= 840$
View full question & answer→MCQ 1271 Mark
The sum of the integers from $1$ to $100$ which are not divisible by $3$ or $5$ is:
- A
$2489$
- B
$4735$
- ✓
$2317$
- D
$2632$
AnswerCorrect option: C. $2317$
View full question & answer→MCQ 1281 Mark
The consecutive digits of a three digit number are in $GP.$ If the middle digit be increased by $2$, then they form an $AP.$ If $792$ is subtracted from this, then we get the number constituting of same three digits but in reverse order. Then, number is divisible by:
View full question & answer→MCQ 1291 Mark
In numbers from $1$ to $100$ the digit $"0"$ appears $...........$ times:
Answer$10, 20, 30, 40, 50, 60, 70, 80, 90, 100$
Thus the digit $0$ appears $11$ times.
View full question & answer→MCQ 1301 Mark
Concentric circles of radii $1, 2, 3, … , 100 \ cm$ are drawn. The interior of the smallest circle is coloured red and the angular regions are coloured alternately green and red, so that no two adjacent regions are of the same colour. Then, the total area of the green regions in $\text{sq \ cm}$ is equal to:
- A
$1000\pi$
- ✓
$5050\pi$
- C
$4950\pi$
- D
$5151\pi$
AnswerCorrect option: B. $5050\pi$
View full question & answer→MCQ 1311 Mark
$\sum\limits^{4}_{\text{i = 1}}2\text{n}+3=............$
Answer$a_1 = 2 \times 1 + 3 = 5,$
$a_2 = 2 \times 2 + 3 = 7,$
$a_3 = 2 \times 3 + 3 = 9,$
$a_4= 2 \times 4 + 3 = 11.$
Sum $= 5 + 7 + 9 + 11 = 32.$
View full question & answer→MCQ 1321 Mark
If $\big(4^3\big)\big(4^6\big)\big(4^9\big)\big(4^{12}\big)\ \dots\big(4^{3\text{x}}\big)=(0.0625)^{-54},$ the value of x is:
Answer
Solution: (B) 8
$\big(4^3\big)\big(4^6\big)\big(4^9\big)\big(4^{12}\big)\ \dots\big(4^{3\text{x}}\big)=(0.0625)^{-54}$
$\Rightarrow4^{(3+6+9+12+\ \dots+3\text{x})}=\Big(\frac{625}{10000}\Big)^{-54}$
$\Rightarrow4^{3(1+2+3+4+\dots+\text{x})}=\Big(\frac{1}{16}\Big)^{-54}$
$\Rightarrow4^{3\big(\frac{\text{x}(\text{x}+1)}{2}\big)}=\Big(\frac{1}{16}\Big)^{-54}$
$\Rightarrow4^{3\big(\frac{\text{x}(\text{x}+1)}{2}\big)}=\Big(4^{-2}\Big)^{-54}$
Comparing both the sides:
$\Rightarrow3\Big(\frac{\text{x}(\text{x}+1)}{2}\Big)=108$
$\Rightarrow\text{x}(\text{x}+1)=72$
$\Rightarrow\text{x}^2+\text{x}-72=0$
$\Rightarrow\text{x}^2+9\text{x}-8\text{x}-72=0$
$\Rightarrow\text{x}(\text{x}+9)-8(\text{x}+9)=0$
$\Rightarrow(\text{x}+9)(\text{x}-8)=0$
$\Rightarrow\text{x}=8,-9$
$\Rightarrow\text{x}=8$ $[\because\text{ x}\text{ is psitive}]$
View full question & answer→MCQ 1331 Mark
Which term of $G.P. 25, 125, 625, ………….$ is $390625?$
AnswerIn the given $G.P.$, In the given $G.P., a=25$ and
$\text{r}=\frac{125}{25}=5$
Given, $a_n= 390625$
$\Rightarrow ar^{n-1}= 390625$
$\Rightarrow 25 \times 5^{n-1} = 390625$
$\Rightarrow 5^\text{n-1}=\frac{390625}{25}=15625=5^6$
$\Rightarrow n-1 = 6$
$\Rightarrow n=7.$
View full question & answer→MCQ 1341 Mark
If $\sum\text{n}=210,$ then $\sum\text{n}^2=$
AnswerCorrect option: A. $2870$
Given,
$\sum\text{n}=210$
$\Rightarrow\text{n}\Big(\frac{\text{n}+1}{2}\Big)=210$
$\Rightarrow\text{n}^2+\text{n}-420=0$
$\Rightarrow(\text{n}-20)(\text{n}+21)=0$
$\Rightarrow\text{n}=20$
$(\because\ \text{n}>0)$
Now,
$\sum\text{n}^2=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
$\Rightarrow\frac{\text{n}(\text{n}+1)}{2}\times\frac{(2\text{n}+1)}{3}$
$\Rightarrow(210)\times\Big(\frac{41}{3}\Big)$
$\Rightarrow(70)\times(41)$
$\Rightarrow2870$
View full question & answer→MCQ 1351 Mark
Complete $2, 3, 5, 7, ..........$
AnswerSince $2, 3, 5$ and $7$ all are consecutive prime numbers
so, it is a sequence of prime numbers.
Prime number next to $7$ is $11.$
So, $2, 3, 5, 7, 11.$
View full question & answer→MCQ 1361 Mark
Find the sum up to $7^{th}$ term of series $2+3+5+8+12+…………$
Answer$S_n = 2+3+5+8+12+……………+ a_n$
$S_n = 2+3+5+8+12+ ……. + a_{n-1}+ a_n$
Subtracting we get, $0 = 2+1+2+3+4+………… – a_n$
$\Rightarrow\text{a}^\text{n}=2+1+2+3+4+...........+(\text{n}+1)$
$=2+(\text{n}+1)\frac{\text{n}}{2}=\Big(\frac{1}{2}\Big)\text{(n}^2-\text{n}+4)\text{n}^\text{th}\text{terms}\text{ is}\Big(\frac{1}{2}\Big)(\text{n}^2-\text{n}+4$ so $\text{ a}_\text{k} $
$=2+(\text{n}+1)\frac{\text{n}}{2}$
$=\Big(\frac{1}{2}\Big)(\text{k}^2-\text{k+4)}$
Taking summation from $k=1$ to $k=n$ on both sides, we get
$\sum\limits^\text{n}_\text{i}=0\text{ a}_\text{k}=\Big(\frac{1}{2}\Big)\sum\limits^\text{n}_\text{i}=0\text{ k}^2-\Big(\frac{1}{2}\Big)\sum\limits^\text{n}_\text{i}=0\text{ k}+2\text{n}$
$=\frac{\text{n}(\text{n+1)}(2\text{n}+1)}{(2\times6)}$
Here $,\text{n}=7.$
so, $\sum\limits^\text{n}_\text{i}=0\text{ a}\text{k}=\frac{(7\times8\times15)}{12}-\frac{(7\times8)}{4}+2\times7=70.$
View full question & answer→MCQ 1371 Mark
If $a_n = 4n + 6$, find $15^{th}$ term of the sequence.
Answer$a_n = 4n + 6$ and $n = 15$
$\Rightarrow a_{15}= 4 \times 15 + 6$
$= 60 + 6$
$= 66.$
View full question & answer→MCQ 1381 Mark
Insert $4$ numbers between $2$ and $22$ such that the resulting sequence is an $A.P.$
- A
$4, 8, 12, 16$
- B
$5, 9, 13, 17$
- C
$4, 10, 15, 19$
- ✓
$6, 10, 14, 18$
AnswerCorrect option: D. $6, 10, 14, 18$
Let $A.P.$ be $2, A_1, A_2, A_3, A_4, 22$.
$\Rightarrow a=2$ and $a_6=a+5 d=22$
$\Rightarrow 2+5 \times d=22$
$\Rightarrow d=4$
$ A_1=a_2=a+d=2+4=6 $
$ A_2=A_1+d=6+4=10 $
$ A_3=10+4=14 $
$ A_4=14+4=18$
View full question & answer→MCQ 1391 Mark
If $a_1, a_2, \ldots a_n$ are in $HP,$ then the expression $a_1 a_2+a_2 a_3+\ldots a_n-1$ is equal to:
- A
$(n-1) (a_1 – a_n )$
- B
$na_1 a_n$
- ✓
$(n-1) a_1 a_n$
- D
$n (a_1- a_n)$
AnswerCorrect option: C. $(n-1) a_1 a_n$
View full question & answer→MCQ 1401 Mark
If $p, q$ be two $A.M.'s$ and $G$ be one $G.M.$ between two numbers, then $G^2 =$
- ✓
$(2\text{p}-\text{q})(\text{p}-2\text{q})$
- B
$(2\text{p}-\text{q})(2\text{q}-\text{p})$
- C
$(2\text{p}-\text{q})(\text{p}+2\text{q})$
- D
AnswerCorrect option: A. $(2\text{p}-\text{q})(\text{p}-2\text{q})$
Let the two numbers be $a$ and $b.$
$a, p, q$ and $b$ are in $A.P.$
$\therefore\text{ p}-\text{a}=\text{q}-\text{q}=\text{b}-\text{q}$
$\Rightarrow\text{ p}-\text{a}=\text{q}-\text{p}$ and $\text{ q}-\text{p}=\text{b}-\text{q}$
$\Rightarrow\text{ a}=2\text{p}-\text{q}$ and $\text{ b}=2\text{q}-\text{p}\cdots(\text{i})$
Also, $a, G$ and $b$ are in $G.P.$
$\therefore\text{G}^2=\text{ab}$
$\Rightarrow\text{G}^2=(2\text{p}-\text{q})(2\text{q}-\text{p})$
View full question & answer→MCQ 1411 Mark
A man saves $Rs. 200$ in each of the first three months of his service. In each of the subsequent months his saving increases by $Rs. 40$ more than the saving of immediately previous month. His total saving from the start of service will be $Rs. 11040$ after:
- A
$19$ months
- B
$20$ months
- ✓
$21$ months
- D
$18$ months
AnswerCorrect option: C. $21$ months
View full question & answer→MCQ 1421 Mark
The sum to $n$ terms of the series $\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\ ....\text{ is}:$
AnswerCorrect option: D. $\frac{1}{2}\big\{\sqrt{2\text{n}+1}-1\big\}$
Let $T_n$ be the $n^{th}$ term of the given series.
Thus, we have
$\text{T}_\text{n}=\frac{1}{\sqrt{2\text{n}-1}+\sqrt{2\text{n}+1}}$
$=\frac{\sqrt{2\text{n}+1}-\sqrt{2\text{n}-1}}{2}$
Now,
Let $S_n$ be the sum $n$ terms of the given series.
Thus, we have
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}$
$=\sum\limits^{\text{n}}_{\text{k}=1}\bigg(\frac{\sqrt{2\text{k}+1}-\sqrt{2\text{k}-1}}{2}\bigg)$
$=\frac{1}{2}\sum\limits^{\text{n}}_{\text{k}=1}\big(\sqrt{2\text{k}+1}-\sqrt{2\text{k}-1}\big)$
$=\frac{1}{2}\Big[\big(\sqrt{3}-\sqrt{1}\big)+\big(\sqrt{5}-\sqrt{3}\big)+\big(\sqrt{7}-\sqrt{5}\big)+\ ...\ +\big(\sqrt{2\text{n}+1}-\sqrt{2\text{n}-1}\big)\Big]$
$=\frac{1}{2}\Big\{(-1)+\sqrt{2\text{n}+1}\Big\}$
$=\frac{1}{2}\big\{\sqrt{2\text{n}+1}-1\big\}$
View full question & answer→MCQ 1431 Mark
Find the sum of series up to $6^{th}$ term whose $n^{th}$ term is given by $n^2 + 3^n$.
- A
$91$
- B
$1284$
- ✓
$1183$
- D
$1092$
AnswerCorrect option: C. $1183$
Given, $n^{th}$ term is $n^2+ 3^n$
So, $a_k= k^2 + 3^k$
Taking summation from $k=1$ to $k=n$ on both sides, we get
$\sum\limits^\text{n}_\text{i}=0\text{ a}\text{k}=\sum\limits^\text{n}_\text{i}=0\text{ k}^2+\sum\limits^\text{n}_\text{i}=0 3 ^\text{k}$
$\sum\limits^\text{n}_\text{i}=0\text{ k}^2=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
$\sum\limits^\text{n}_\text{i}=0 3^\text{ k}=\frac{3\times(3^\text{n}-1)}{(3-1)}=\Big(\frac{3}{2}\Big)(3^\text{n}-1)$
$\sum\limits^\text{n}_\text{i}=0\text{ a}_\text{k}=\frac{\text{n}(\text{n+1}(2\text{n+1)}}{6}+\Big(\frac{3}{2}\Big)(3^\text{n}-1) $
Sum up to $6^{th}$ term
$=\frac{6\times7\times13}{6}+\Big(\frac{3}{2}\Big)(3^6-1)$
$=91+1092$
$=1183.$
View full question & answer→MCQ 1441 Mark
If two numbers are $2$ and $6$ then find their arithmetic mean.
AnswerExplanation: We know that arithmetic mean of two numbers is given by the average of two numbers
i.e. $A.M.$
$\frac{(2+6)}{2}=\frac{8}{2}=4.$
View full question & answer→MCQ 1451 Mark
Complete $2, 4, 6, 8, ...........$
AnswerSince $2, 4, 6$ and $8$ are even numbers so it is a sequence of even numbers.
Even number next to $8$ is $10.$
So, $2, 4, 6, 8, 10.$
View full question & answer→MCQ 1461 Mark
If the sum of first two terms of an infinite $G.P.$ is $1$ and every term is twice the sum of all the successive terms, then its first term is:
- A
$\frac13$
- B
$\frac23$
- C
$\frac14$
- ✓
$\frac34.$
AnswerCorrect option: D. $\frac34.$
Let the terms of the $G.P.$ be $\text{a},\text{a}_2,\text{a}_3,\text{a}_4.\text{a}_5,\ \dots,\infty.$
And, let the common ratio be $r.$
Now, $\text{a}+\text{a}_2=1$
$\therefore\text{a}+\text{ar}=1\dots(\text{i})$
Also, $\text{a}=2(\text{a}_2+\text{a}_3+\text{a}_4+\text{a}_5+\dots\infty)$
$\Rightarrow\text{a}=2\big(\text{ar}+\text{ar}^2+\text{ar}^3+\text{ar}^4+\ \dots\infty\big)$
$\Rightarrow\text{a}=2\Big(\frac{\text{ar}}{1-\text{r}}\Big)$
$\Rightarrow1-\text{r}=2\text{r}$
$\Rightarrow3\text{r}=1$
$\Rightarrow\text{r}=\frac13$
Putting the value of $r$ in $(i):$
$\text{a}+\frac{a}{3}=1$
$\Rightarrow\frac{4\text{a}}{3}=1$
$\Rightarrow4\text{a}=3$
$\Rightarrow\text{a}=\frac34$
View full question & answer→MCQ 1471 Mark
If in an $A.P.,$ first term is $20,$ common difference is $2$ and nth term is $42,$ then find sum up to $n$ terms.
AnswerWe know, $a = 20, d = 2, a_n= 42.$
$a + (n - 1) d = 42$
$\Rightarrow 20 + 2(n - 1) = 42$
$\Rightarrow 2(n - 1) = 42 - 20 = 22$
$\Rightarrow n - 1 = 11$
$\Rightarrow n = 12$
$\text{s}_\text{a}=\frac{\text{n}}{2}(a+I)$
$\Rightarrow\text{s}_{\text{12}=\frac{12}{2}}(20+42)=6\times62=372$
View full question & answer→MCQ 1481 Mark
Choose the correct answer. If the sum of $n$ terms of an $A.P.$ is given by $S_n = 3n + 2n^2$, then the common difference of the $A.P.$ is:
AnswerGiven that:
$ S_n=3 n+2 n^2 $
$ S_1=3(1)+2(1) 2=5 $
$ S_2=3(2)+2(4)=14 $
$ S_1=a_1=5 $
$ S_2-S_1=a_2=14-5=9$
$\therefore$ Common difference $\mathrm{d}=\mathrm{a}_2-\mathrm{a}_1=9-5=4$
Hence, the correct option is $(d).$
View full question & answer→MCQ 1491 Mark
If $|x| < 1,$ then the sum of the series $1 + 2x + 3 \times 2 + 4 \times 3…\infty $ will be:
- A
$1/1 - x$
- B
$1/1 + x$
- C
$1(1 + x^2)$
- ✓
$1(1 - x^2)$
AnswerCorrect option: D. $1(1 - x^2)$
View full question & answer→MCQ 1501 Mark
If $S$ be the sum, $P$ the product and $R$ be the sum of the reciprocals of n terms of a $\text{G.P.}$ then $P^2$ is equal to:
- A
$\frac{\text{S}}{\text{R}}$
- B
$\frac{\text{R}}{\text{S}}$
- C
$\Big(\frac{\text{R}}{\text{S}}\Big)^\text{n}$
- ✓
$\Big(\frac{\text{S}}{\text{R}}\Big)^\text{n}.$
AnswerCorrect option: D. $\Big(\frac{\text{S}}{\text{R}}\Big)^\text{n}.$
Sum of $n$ terms of the $\text{G.P.}, \text{S}=\frac{\text{a}(\text{r}^{\text{n}}-1)}{(\text{r}-1)}$
Product of $n$ terms of the $\text{G.P.}, \text{P}=\text{a}^{\text{n}}\text{r}^{\big[\frac{\text{n}(\text{n}-1)}{2}\big]}$
Sum of the reciprocals of $n$ terms of the $\text{G.P.}, \text{R}=\frac{\Big[\frac{1}{\text{r}^\text{n}}-1\Big]}{\text{a}\big(\frac{1}{\text{r}}-1\big)}=\frac{(\text{r}^{\text{n}}-1)}{\text{ar}^{(\text{n}-1)}(\text{r}-1)}$
$\therefore\text{P}^2=\bigg\{\text{a}^2\text{r}^\frac{2(\text{n}-1)}{2}\bigg\}^\text{n}$
$\Rightarrow\text{P}^2=\Bigg\{\frac{\frac{\text{a}(\text{r}^\text{n}-1)}{(\text{r}-1)}}{\frac{(\text{r}^\text{n}-1)}{\text{ar}^{(\text{n}-1)(\text{r}-1)}}}\Bigg\}^\text{n}$
$\Rightarrow\text{P}^2=\Big\{\frac{\text{S}}{\text{R}}\Big\}^\text{n}$
Let the first term of the $\text{G.P.}$ be a and the common ratio be $r.$
Sum of $n$ terms, $\text{S}=\frac{\text{a}(\text{r}^\text{n}-1)}{\text{r}-1}$
Product of the $\text{G.P.,}$ $\text{P}=\text{a}^{\text{n}}\text{r}^{\frac{\text{n}(\text{n}+1)}{2}}$
Sum of the reciprocals of $n$ terms,
$\Rightarrow\text{R}=\frac{\big(\frac{1}{\text{r}^{\text{n}-1}}\big)}{\text{a}\big(\frac{1}{\text{r}-1}\big)}$
$=\frac{\big(\frac{1-\text{r}^{\text{n}}}{\text{r}^\text{n}}\big)}{\text{a}\big(\frac{1}{\text{r}-1}\big)}$
$\text{P}^2=\bigg\{\text{a}^2\text{r}^{\frac{(\text{n}+1)}{2}}\bigg\}^{\text{n}}$
$\text{P}^2=\begin{Bmatrix} \frac{\frac{\text{a}(\text{r}^\text{n}-1)}{\text{r}-1}}{\frac{\Big(\frac{1-\text{r}^\text{n}}{\text{r}^\text{n}}\Big)}{\text{a}\Big(\frac{1-\text{r}}{\text{r}}\Big)}}\end{Bmatrix}$
$=\Big\{\frac{\text{S}}{\text{R}}\Big\}^\text{n}$
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