MCQ 2011 Mark
Let $a_1=8, a_2, a_3, \ldots a_n$ be an $A.P.$ If the sum of its first four terms is $50$ and the sum of its last four terms is $170$ , then the product of its middle two terms is
Answerc
$a_1+a_2+a_3+a_4=50$
$\Rightarrow 32+6 d=50$
$\Rightarrow d=3$
and, $a_{n-3}+a_{n-2}+a_{n-1}+a_n=170$
$\Rightarrow 32+(4 n -10) \cdot 3=170$
$\Rightarrow n =14$
$a _7=26, a _8=29$
$\Rightarrow a _7 \cdot a _8=754$
View full question & answer→MCQ 2021 Mark
The $8^{\text {th }}$ common term of the series $S _1=3+7+11+15+19+\ldots . .$ ; $S _2=1+6+11+16+21+\ldots .$ is $.......$.
Answerb
$T _8=11+(8-1) \times 20$
$=11+140=151$
View full question & answer→MCQ 2031 Mark
Let $s _1, s _2, s _3, \ldots \ldots, s _{10}$ respectively be the sum to 12 terms of 10 A.P.s whose first terms are $1,2,3, \ldots, 10$ and the common differences are $1,3,5, \ldots, 19$ respectively. Then $\sum \limits_{i=1}^{10} s _{ i }$ is equal to
- A
$7380$
- B
$7220$
- C
$7360$
- ✓
$7260$
AnswerCorrect option: D. $7260$
d
$S _{ k }=6(2 k +(11)(2 k -1))$
$S _{ k }=6(2 k +22 k -11)$
$S _{ k }=144 k -66$
$\sum \limits_1^{10} S _{ k }=144 \sum \limits_{ k =1}^{10} k -66 \times 10$
View full question & answer→MCQ 2041 Mark
Let the digits $a, b, c$ be in $A.P.$ Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in $A.P.$ at least once. How many such numbers can be formed?
- A
$1261$
- B
$1262$
- C
$1263$
- ✓
$1260$
AnswerCorrect option: D. $1260$
d
$\frac{{ }^7 C _1 \times 2 \times 6 !}{2 ! 2 ! 2 !}=1260$
View full question & answer→MCQ 2051 Mark
Let $x _1, x _2 \ldots ., x _{100}$ be in an arithmetic progression, with $x _1=2$ and their mean equal to $200$ . If $y_i=i\left(x_i-i\right), 1 \leq i \leq 100$, then the mean of $y _1, y _2$, $y _{100}$ is
- A
$10101.50$
- B
$10051.50$
- ✓
$10049.50$
- D
$10100$
AnswerCorrect option: C. $10049.50$
c
$\text { Mean }=200$
$\Rightarrow \frac{\frac{100}{2}(2 \times 2+99 d)}{100}=200$
$\Rightarrow 4+99 d =400$
$\Rightarrow d=4$
$y_i=i(x i-i)$
$=i(2+(i-1) 4-i)=3 i^2-2 i$
$\text { Mean }=\frac{\sum y_i}{100}$
$=\frac{1}{100} \sum \limits_{i=1}^{100} 3 i^2-2 i$
$=\frac{1}{100}\left\{\frac{3 \times 100 \times 101 \times 201}{6}-\frac{2 \times 100 \times 101}{2}\right\}$
$=101\left\{\frac{201}{2}-1\right\}=101 \times 99.5$
$=10049 \cdot 50$
View full question & answer→MCQ 2061 Mark
The sum of all those terms, of the anithmetic progression $3,8,13, \ldots \ldots .373$, which are not divisible by $3$,is equal to $.......$.
- A
$9524$
- B
$9523$
- C
$9522$
- ✓
$9525$
AnswerCorrect option: D. $9525$
d
$\text { Required sum }=(3+8+13+18+\ldots \ldots \ldots+373)$
$-(3+18+33+\ldots \ldots+363)$
$=\frac{75}{2}(3+373)-\frac{25}{2}(3-363)$
$=75 \times 188-25 \times 183$
$=9525$
View full question & answer→MCQ 2071 Mark
The sum of the common terms of the following three arithmetic progressions.
$3,7,11,15,...................,399$
$2,5,8,11,............,359$ and
$2,7,12,17,...........,197$, is equal to $................$.
Answerb
$3,7,11,15, \ldots \ldots \ldots \ldots, 399 \quad d_1=4$
$2,5,8,11, \ldots \ldots \ldots \ldots ., 359 \quad d_2=3$
$2,7,12,17, \ldots \ldots ., 197 \quad d_3=5$
$\operatorname{LCM}\left(d_1, d_2, d_3\right)=60$
Common terms are $47, 107, 167$
$Sum =321$
View full question & answer→MCQ 2081 Mark
Let $a_1, a_2, \ldots \ldots, a_n$ be in A.P. If $a_5=2 a_3$ and $a_{11}=18$, then $12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots . \cdot \frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$ is equal to $..........$.
Answera
$2 a_7=a_s(\text { given })$
$2\left(a_1+6 d\right)=a_1+4 d$
$a_1+8 d=0$
$a_1+10 d=18$
$\text { By }(1) \text { and }(2) \text { we get } a_1=-72, d=9$
$a_{18}=a_1+17 d=-72+153=81$
$a_{10}=a_1+9 d=9$
$12\left(\frac{\sqrt{a_{11}}-\sqrt{a_{10}}}{d}+\frac{\sqrt{a_{12}}-\sqrt{a_{11}}}{d}+\ldots . . \frac{\sqrt{a_{18}}-\sqrt{a_{17}}}{d}\right)$
$12\left(\frac{\sqrt{a_{18}}-\sqrt{a_{10}}}{d}\right)=\frac{12(9-3)}{9}=\frac{12 \times 6}{6}=8$
View full question & answer→MCQ 2091 Mark
For three positive integers $p , q , r , x ^{ pq p ^2}= y ^{ qr }= z ^{ p ^2 r }$ and $r=p q+1$ such that $3,3 \log _y x, 3 \log _z y, 7 \log _x z$ are in A.P. with common difference $\frac{1}{2}$. Then $r - p - q$ is equal to
Answera
$pq ^2=\log _{ x } \lambda$
$qr =\log _{ y } \lambda$
$p ^2 r =\log _{ z } \lambda$
$\log _{ y } x =\frac{ qr }{ pq ^2}=\frac{ r }{ pq } \ldots(1)$
$\log _{ x } z =\frac{ pq ^2}{ p ^2 r }=\frac{ q ^2}{ pr } \ldots(2)$
$\log _{ z } y =\frac{ p ^2 r }{ qr }=\frac{ p ^2}{ q } \ldots \ldots(3)$
$3, \frac{3 r }{ pq }, \frac{3 p ^2}{ q }, \frac{7 q ^2}{ pr } \text { in A.P }$
$\frac{3 r }{ pq }-3=\frac{1}{2}$
$r =\frac{7}{6} pq.....(4)$
$r = pq +1$
$pq =6 \ldots(5)$
$r =7 \ldots \ldots(6)$
$\frac{3 p ^2}{ q }=4$
After solving $p =2$ and $q =3$
View full question & answer→MCQ 2101 Mark
Let $a _1, a _2, a _3, \ldots$ be a $G.P.$ of increasing positive numbers. Let the sum of its $6^{\text {th }}$ and $8^{\text {th }}$ terms be $2$ and the product of its $3^{\text {rd }}$ and $5^{\text {th }}$ terms be $\frac{1}{9}$. Then $6\left( a _2+\right.$ $\left.a_4\right)\left(a_4+a_6\right)$ is equal to
- A
$2 \sqrt{2}$
- B
$2$
- C
$3 \sqrt{3}$
- ✓
$3$
Answerd
$a r^5+a r^7=2$
$\left(a r^2\right)\left(a r^4\right)=\frac{1}{9}$
$a^2 r^6=\frac{1}{9}$
Now, $r > 0$
$\operatorname{ar}^5\left(1+r^2\right)=2$
Now, $ar ^3=\frac{1}{3}$ or $-\frac{1}{3}$ (rejected)
$r^2=2$
$r=\sqrt{2}$
$a=\frac{1}{6 \sqrt{2}}$
Now, $6\left(a_2+a_4\right)\left(a_4+a_6\right)$
$6\left(a r+a r^3\right)\left(a r^3+a r^5\right)$
$6 a^2 r^4\left(1+r^2\right)$
$6\left(\frac{1}{36.2}\right)(4)(9)=3$
View full question & answer→MCQ 2111 Mark
Let the positive numbers $a _1, a _2, a _3, a _4$ and $a _5$ be in a G.P. Let their mean and variance be $\frac{31}{10}$ and $\frac{ m }{ n }$ respectively, where $m$ and $n$ are co-prime. If the mean of their reciprocals is $\frac{31}{40}$ and $a_3+a_4+a_5=14$, then $m + n$ is equal to $.........$.
Answerd
Let $\frac{a}{r}, \frac{a}{r}, a, a r, a r^2$
$\text { Given } \frac{a}{r^2}+\frac{a}{r}+a+a r+a r^2=5 \times \frac{31}{10}$
$\text { And } \frac{r^2}{a}+\frac{r}{a}+\frac{1}{a}+\frac{1}{a r}+\frac{1}{a r^2}=5 \times \frac{31}{40}$
$(1) \div(2) a^2=4 \Rightarrow a=2 \quad \therefore r+\frac{1}{r}=5 / 2 \quad(a \neq-2)$
$\Rightarrow r=2$
$\therefore \text { Now } \frac{1}{2}, 1,2.4,8$
$\therefore \sigma^2=\frac{\sum x^2}{N}-\left(\frac{\sum x }{ N }\right)^2$
$=\frac{186}{25}=\frac{ M }{N} \Rightarrow 211= m + n$
View full question & answer→MCQ 2121 Mark
Let the first term $a$ and the common ratio $r$ of a geometric progression be positive integers. If the sum of its squares of first three terms is $33033$, then the sum of these three terms is equal to
Answera
$\Rightarrow a^2+a^2 r ^2+a^2 r ^4=33033$
$\Rightarrow a^2\left( r ^4+ r ^2+1\right)=3 \times 7 \times 11^2 \times 13 \Rightarrow a=11$
$\Rightarrow r ^4+r^2+1=273 \quad \Rightarrow r^4+r^2-272=0$
$\Rightarrow\left(r^2+17\right)\left(r^2-16\right)=0 \Rightarrow r^2=16 \Rightarrow r = \pm 4$
$t_1+t_2+t_3=a+a r+a r^2=11+44+176=231$
View full question & answer→MCQ 2131 Mark
If the sum and product of four positive consecutive terms of a $G.P.$, are $126$ and $1296$, respectively, then the sum of common ratios of all such $GPs$ is $.........$.
- ✓
$7$
- B
$\frac{9}{2}$
- C
$3$
- D
$14$
Answera
$a, a r, a r^2, a r^3(a, r > 0)$
$a^4 r^6=1296$
$a^2 r^3=36$
$a=\frac{6}{r^{3 / 2}}$
$a+a r+a r^2+a r^3=126$
$\frac{1}{r^{3 / 2}}+\frac{r}{r^{3 / 2}}+\frac{r^2}{r^{3 / 2}}+\frac{r^3}{r^{3 / 2}}=\frac{126}{6}=21$
$\left(r^{-3 / 2}+r^{3 / 2}\right)+\left(r^{1 / 2}+r^{-1 / 2}\right)=21$
$r^{1 / 2}+r^{-1 / 2}=A$
$r^{-3 / 2}+r^{3 / 2}+3 A = A ^3$
$A ^3-3 A + A =21$
$A ^3-2 A =21$
$A =3$
$\sqrt{ r }+\frac{1}{\sqrt{r}}=3$
$r +1=3 \sqrt{ r }$
$r^2+2 r+1=9 r$
$r^2-7 r+1=0$
View full question & answer→MCQ 2141 Mark
Let $\left\{a_k\right\}$ and $\left\{b_k\right\}, k \in N$, be two G.P.s with common ratio $r_1$ and $r_2$ respectively such that $a_1=b_1=4$ and $r_1 < r_2$. Let $c_k=a_k+k, \in N$. If $c_2=5$ and $c_3=13 / 4$ then $\sum \limits_{k=1}^{\infty} c_k - \left(12 a _6+8 b _4\right)$ is equal to
Answera
Given that
$c_k=a_k+b_k$ and
also
$a _2=4 r _1$ $\quad a _3=4 r _1{ }^2$
$b _2=4 r _2$ $\quad b _3=4 r _2{ }^2$
Now $c_2=a_2+b_2=5$ and $c_3=a_3+b_3=\frac{13}{4}$
$\Rightarrow r_1+r_2=\frac{5}{4}$ and $r_1^2+r_2^2=\frac{13}{16}$
Hence $r_1 r_2=\frac{3}{8}$ which gives $r_1=\frac{1}{2} \quad \& \quad r_2=\frac{3}{4}$
$\sum \limits_{ k =1}^{\infty} c _{ k }-\left(12 a _6+8 b _4\right)$
$=\frac{4}{1-r_1}+\frac{4}{1-r_2}-\left(\frac{48}{32}+\frac{27}{2}\right)$
$=24-15=9$
View full question & answer→MCQ 2151 Mark
Let $a_1, a_2, a_3, \ldots$. be a $GP$ of increasing positive numbers. If the product of fourth and sixth terms is $9$ and the sum of fifth and seventh terms is $24 ,$ then $a_1 a_9+a_2 a_4 a_9+a_5+a_7$ is equal to $.........$.
Answerc
$a_4 \cdot a_6=9 \Rightarrow\left(a_5\right)^2=9 \Rightarrow a_5=3$
$a_5+a_7=24 \Rightarrow a_5+a_5 r^2=24 \Rightarrow\left(1+r^2\right)=8 \Rightarrow r=\sqrt{7}$ $\Rightarrow a=\frac{3}{49}$
$\Rightarrow a_1 a_9+a_2 a_4 a_9+a_5+a_7=9+27+3+21=60$
View full question & answer→MCQ 2161 Mark
The $4^{\text {tht }}$ term of $GP$ is $500$ and its common ratio is $\frac{1}{m}, m \in N$. Let $S_n$ denote the sum of the first $n$ terms of this GP. If $S_6 > S_5+1$ and $S_7 < S_6+\frac{1}{2}$, then the number of possible values of $m$ is $..........$
Answerc
$T_4=500 \quad$ where $a=$ first term,
$r =$ common ratio $=\frac{1}{ m }, m \in N$
$a r^3=500$
$\frac{a}{m^3}=500$
$S_n-S_{n-1}=a r^{n-1}$
$S _6 > S _5+1 \quad$ and $S _7- S _6 < \frac{1}{2}$
$S _6- S _5 > 1 \quad \frac{ a }{ m ^6} < \frac{1}{2}$
$ar ^5 > 1 \quad m ^3 > 10^3$
$\frac{500}{ m ^2} > 1 \quad m > 10$
$m ^2 < 500$
From $(1)$ and $(2)$
$m =11,12,13 \ldots \ldots \ldots \ldots ., 22$
So number of possible values of $m$ is $12$
View full question & answer→MCQ 2171 Mark
Let $A_1$ and $A_2$ be two arithmetic means and $G_1, G_2$, $G _3$ be three geometric means of two distinct positive numbers. The $G _1^4+ G _2^4+ G _3^4+ G _1^2 G _3^2$ is equal to
- A
$2\left( A _1+ A _2\right) G _1 G _3$
- ✓
$\left(A_1+A_2\right)^2 G_1 G_3$
- C
$\left( A _1+ A _2\right) G _1^2 G _3^2$
- D
$2\left( A _1+ A _2\right) G _1^2 G _3^2$
AnswerCorrect option: B. $\left(A_1+A_2\right)^2 G_1 G_3$
b
$a , A _1, A _2, b$ are in A.P.
$d =\frac{b-a}{3} ; A_1=a+\frac{b-a}{3}=\frac{2 a+b}{3}$
$A_2=\frac{a+2 b}{3}$
$A_1+A_2=a+b$
$a, G_1, G_2, G_3, b \text { are in G.P. }$
$r=\left(\frac{b}{a}\right)^{\frac{1}{4}}$
$G_1=\left(a^3 b\right)^{\frac{1}{4}}$
$G_2=\left(a^2 b^2\right)^{\frac{1}{4}}$
$G_3=\left(a^3\right)^{\frac{1}{4}}$
$G_1^4+G_2^4+G_3^4+G_1^2 G_3^2=$
$a^3 b+a^2 b^2+a b^3+\left(a^3 b\right)^{\frac{1}{2}} \cdot\left(a b^3\right)^{\frac{1}{2}}$
$=a^3 b+a^2 b^2+a b^3+a^2 \cdot b^2$
$=a b\left(a^2+2 a b+b^2\right)$
$=a b(a+b)^2$
$=G_1 \cdot G_3 \cdot\left(A_1+A_2\right)^2$
View full question & answer→MCQ 2181 Mark
Let $a , b , c$ and $d$ be positive real numbers such that $a+b+c+d=11$. If the maximum value of $a^5 b^3 c^2 d$ is $3750 \beta$, then the value of $\beta$ is
Answera
$\frac{5\left(\frac{ a }{5}\right)+3\left(\frac{ b }{3}\right)+2\left(\frac{ c }{2}\right)+ d }{11} \geq\left(\frac{ a ^5 b ^3 c ^2 d }{5^5 3^3 2^2}\right)^{1 / 11}$
$1 \geq\left(\frac{a^5 b^3 c^2 d}{5^5 3^3 2^2}\right)^{1 / 11}$
$\beta=90$
View full question & answer→MCQ 2191 Mark
Let $0 < z < y < x$ be three real numbers such that $\frac{1}{ x }, \frac{1}{ y }, \frac{1}{ z }$ are in an arithmetic progression and $x$, $\sqrt{2} y, z$ are in a geometric progression. If $x y+y z$ $+z x=\frac{3}{\sqrt{2}} x y z$, then $3(x+y+z)^2$ is equal to $............$.
Answera
$\frac{2}{y}=\frac{1}{x}+\frac{1}{z}$
$2 y^2=x z$
$\frac{2}{y}=\frac{x+z}{x z}=\frac{x+z}{2 y^2}$
$x+z=4 y$
$x y+y z+z x=\frac{3}{\sqrt{2}} x y z$
$y(x+z)+z x=\frac{3}{\sqrt{2}} x z \cdot y$
$4 y^2+2 y^2=\frac{3}{\sqrt{2}} y \cdot 2 y^2$
$6 y^2=3 \sqrt{2} y^3$
$y=\sqrt{2}$
$x+y+z=5 y=5 \sqrt{2}$
$3(x+y+z)^2=3 \times 50=150$
View full question & answer→MCQ 2201 Mark
Let $a, b, c > 1, a^3, b^3$ and $c^3$ be in $A.P.$, and $\log _a b$, $\log _c a$ and $\log _b c$ be in G.P. If the sum of first $20$ terms of an $A.P.$, whose first term is $\frac{a+4 b+c}{3}$ and the common difference is $\frac{a-8 b+c}{10}$ is $-444$, then abc is equal to
- A
$343$
- ✓
$216$
- C
$\frac{343}{8}$
- D
$\frac{125}{8}$
Answerb
As $a ^3, b^3, c^3$ be in $A.P.$ $\rightarrow a^3+c^3=2 b^3$ $\log _{ a }^{ b }, \log _{ c }^{ a }, \log _{ b }^{ c }$ are in $G.P.$
$\therefore \frac{\log b }{\log a } \cdot \frac{\log c}{\log b}=\left(\frac{\log a}{\log c}\right)^2$
$\therefore(\log a)^3=(\log c)^3 \Rightarrow a=c$
From $(1)$ and $(2)$
$a = b = c$
$T _1=\frac{ a +4 b + c }{3}=2 a ; d =\frac{ a -8 b + c }{10}=\frac{-6 a }{10}=\frac{-3}{5} a$
$\therefore S _{20}=\frac{20}{2}\left[4 a +19\left(-\frac{3}{5} a \right)\right]$
$=10\left[\frac{20 a -57 a }{5}\right]$
$=-74 a$
$\therefore-74 a =-444 \Rightarrow a =6$
$\therefore abc =6^3=216$
View full question & answer→MCQ 2211 Mark
For the two positive numbers $a , b$, if $a , b$ and $\frac{1}{18}$ are in a geometric progression, while $\frac{1}{ a }, 10$ and $\frac{1}{ b }$ are in an arithmetic progression, then, $16 a+12 b$ is equal to $.........$.
Answera
$a , b , \frac{1}{18} \rightarrow GP$
$\frac{ a }{18}= b ^2$
$\frac{1}{ a }, 10, \frac{1}{ b } \rightarrow AP$
$\frac{1}{ a }+\frac{1}{ b }=20$
$\Rightarrow a + b =20 ab , \text { from eq. (i) } ; \text { we get }$
$\Rightarrow 18 b ^2+ b =360 b ^3$
$\Rightarrow 360 b ^2-18 b -1=0 \quad\{\because b \neq 0\}$
$\Rightarrow b =\frac{18 \pm \sqrt{324+1440}}{720}$
$\Rightarrow b=\frac{18+\sqrt{1764}}{720} \quad\{\because b > 0\}$
$\Rightarrow b=\frac{1}{12}$
$\Rightarrow a=18 \times \frac{1}{144}=\frac{1}{8}$
Now, $16 a+12 b=16 \times \frac{1}{8}+12 \times \frac{1}{12}=3$
View full question & answer→MCQ 2221 Mark
Suppose $a_1, a_2, 2, a_3, a_4$ be in an arithmeticogeometric progression. If the common ratio of the corresponding geometric progression is $2$ and the sum of all $5$ terms of the arithmetico-geometric progression is $\frac{49}{2}$, then $a_4$ is equal to $...........$.
Answerc
$\frac{(a-2 d)}{4}, \frac{(a-d)}{2}, a, 2(a+d), 4(a+2 d)$
$\left(\frac{1}{4}+\frac{1}{2}+1+6\right) \times 2+(-1+2+8) d=\frac{49}{2}$
$\left(\frac{3}{4}+7\right)+9 d=\frac{49}{2}$
$9 d=\frac{49}{2}-\frac{62}{4}=\frac{98-62}{4}=9$
$d=1$
$\Rightarrow a_4=4(a+2 d)$
$=16$
View full question & answer→MCQ 2231 Mark
If $S _{ n }=4+11+21+34+50+\ldots$ to $n$ terms, then $\frac{1}{60}\left( S _{29}- S _9\right)$ is equal to $.......$.
Answerc
$S _{ n }=4+11+21+34+50+\ldots .+ n \text { terms }$ Difference are in $A.P.$
Let $T_n=a n^2+b n+c$
$T _1= a + b + c =4$
$T _2=4 a +2 b + c =11$
$T _3=9 a +3 b + c =21$
By solving these $3$ equations
$a =\frac{3}{2}, b =\frac{5}{2}, c =0$
So $T_n=\frac{3}{2} n^2+\frac{5}{2} n$
$S _{ n }=\sum T _{ n }$
$=\frac{3}{2} \sum n ^2+\frac{5}{2} \sum n$
$=\frac{3}{2} \frac{ n ( n +1)(2 n +1)}{6}=\frac{5}{2} \frac{( n )( n +1)}{2}$
$=\frac{ n ( n +1)}{4}[2 n +1+5]$
$S _{ n }=\frac{ n ( n +1)}{4}(2 n +6)=\frac{ n ( n +1)( n +3)}{2}$
$\frac{1}{60}\left(\frac{29 \times 30 \times 32}{2}-\frac{9 \times 10 \times 12}{2}\right)=223$
View full question & answer→MCQ 2241 Mark
Let $a_1=b_1=1$ and $a_n=a_{n-1}+(n-1), b_n=b_{n-1}+a_{n-1}$, $\forall n \geq 2$. If $S =\sum \limits_{n=1}^{10} \frac{ b _{ n }}{2^{ n }}$ and $T =\sum \limits_{ n =1}^8 \frac{ n }{2^{ n -1}}$, then $2^7(2 S$ $- T )$ is equal to $........$.
Answera
$\text { As, } S=\frac{b_1}{2}+\frac{b_2}{2^2}+\ldots \ldots .+\frac{b_9}{2^9}+\frac{b_{10}}{2^{10}}$
$\Rightarrow \frac{S}{2}=\quad \frac{b_1}{2^2}+\frac{b_2}{2^3}+\ldots \ldots+\frac{b_9}{2^{10}}+\frac{b_{10}}{2^{11}}$
$\text { subtracting }$
$\Rightarrow \frac{S}{2}=\frac{b_1}{2}+\left(\frac{a_1}{2^2}+\frac{a_2}{2^3} \ldots \ldots+\frac{a_9}{2^{10}}\right)-\frac{b_{10}}{2^{11}}$
$\Rightarrow S=b_1-\frac{b_{10}}{2^{10}}+\left(\frac{a_1}{2}+\frac{a_2}{2^2} \ldots \ldots+\frac{a_9}{2^9}\right)$
$\Rightarrow \frac{S}{2}=\frac{b_1}{2}-\frac{b_{10}}{2^{11}}+\left(\frac{a_1}{2^2}+\frac{a_2}{2^3} \ldots \ldots . .+\frac{a_9}{2^{10}}\right)$
$\text { subtracting }$
$\Rightarrow \frac{ S }{2}=\frac{ b _1}{2}-\frac{ b _{10}}{2^{11}}+\left(\frac{ a _1}{2}-\frac{ a _9}{2^{10}}\right)+\left(\frac{1}{2^2}+\frac{2}{2^3}+\ldots+\frac{8}{2^9}\right.$
$\Rightarrow \frac{ S }{2}=\frac{ a _1+ b _1}{2}-\frac{\left( b _{10}+2 a _9\right)}{2^{11}}+\frac{ T }{4}$
$\Rightarrow 2 S=2\left( a _1+ b _1\right)-\frac{\left( b _{10}+2 a _9\right)}{2^9}+ T$
$\Rightarrow 2^7(2 S - T )=2^8\left( a _1+ b _1\right)-\frac{\left( b _{10}+2 a _9\right)}{4}$
Given $\quad a_n-a_{n-1}=n-1$,
$\therefore \quad a_2-a_1=1$
$\begin{array}{c}a_3-a_2=2 \\ \vdots \\ a_9-a_8=8\end{array}$
$a_9-a_1=1+2+\ldots+8=36$
$a_9=37\left(a_1=1\right)$
$\text { Also, } b_n-b_{n-1}=a_{n-1}$
$\therefore b_{10}-b_1=a_1+a_2+\ldots .+a_9$
$=1+2+4+7+11+16+22+29+37$
$\Rightarrow b_{10}=130\left(A s b_1=1\right)$
$\therefore 2^7(2 S-T)=2^8(1+1)-(130+2 \times 37)$
$2^9-\frac{204}{4}=461$
View full question & answer→MCQ 2251 Mark
The sum to $20$ terms of the series $2.2^2-3^2+2.4^2-5^2+2.6^2-\ldots \ldots$ is equal to $........$.
- A
$1311$
- B
$1312$
- ✓
$1310$
- D
$1313$
AnswerCorrect option: C. $1310$
c
$\left(2^2-3^2+4^2-5^2+20 \text { terms }\right)+ \left(2^2+4^2+\ldots .+10 \text { terms }\right)$
$-(2+3+4+5+\ldots . .+11)+4\left[1+2^2+\ldots \ldots .10^2\right]$
$-\left[\frac{21 \times 22}{2}-1\right]+4 \times \frac{10 \times 11 \times 21}{6}$
$=1-231+14 \times 11 \times 10$
$=1540+1-231$
$=1310$
View full question & answer→MCQ 2261 Mark
Let $\left\langle a _{ n }\right\rangle$ be a sequence such that $a_1+a_2+\ldots+a_n=\frac{n^2+3 n}{(n+1)(n+2)}$. If $28 \sum \limits_{ k =1}^{10} \frac{1}{ a _{ k }}= p _1 p _2 p _3 \ldots p _{ m }$, where $p _1, p _2, \ldots . pm$ are the first $m$ prime numbers, then $m$ is equal to
Answerb
$a_n=S_n-S_{n-1}=\frac{ n ^2+3 n }{( n +1)( n +2)}-\frac{( n -1)( n +2)}{ n ( n +1)}$
$\Rightarrow a _{ n }=\frac{4}{ n ( n +1)( n +2)}$
$\Rightarrow 28 \sum \limits_{ k =1}^{10} \frac{1}{ a _{ k }}=28 \sum \limits_{ k =1}^{10} \frac{ k ( k +1)( k +2)}{4}$
$=\frac{7}{4} \sum \limits_{ k =1}^{10}( k ( k +1)( k +2)( k +3)-( k -1) k ( k +1)( k +2))$
$=\frac{7}{4} \cdot 10 \cdot 11 \cdot 12 \cdot 13=2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$
So $m =6$
View full question & answer→MCQ 2271 Mark
For $k \in N$, if the sum of the series $1+\frac{4}{ k }+\frac{8}{ k ^2}+\frac{13}{ k ^3}+\frac{19}{ k ^4}+\ldots$ is 10 , then the value of $k$ is
Answera
$10=1+\frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\ldots$. upto $\infty$
$9=\frac{4}{ k }+\frac{8}{ k ^2}+\frac{13}{ k ^3}+\frac{19}{ k ^4}+\ldots \text { upto } \infty$
$\frac{9}{ k }=\frac{4}{ k ^2}+\frac{8}{ k ^3}+\frac{13}{ k ^4}+\ldots . \text { upto } \infty$
$S =9\left(1-\frac{1}{ k }\right)=\frac{4}{ k }+\frac{4}{ k ^2}+\frac{5}{ k ^3}+\frac{6}{ k ^4}+\ldots . . \text { upto } \infty$
$\frac{ S }{ k }=\frac{4}{ k ^2}+\frac{4}{ k ^3}+\frac{5}{ k ^4}+\ldots . \text { upto } \infty$
$\left(1-\frac{1}{ k }\right) S =\frac{4}{ k }+\frac{1}{ k ^3}+\frac{1}{ k ^4}+\frac{1}{ k ^3}+\ldots . \infty$
$9\left(1-\frac{1}{ k }\right)^2=\frac{4}{ k }+\frac{\frac{1}{ k ^3}}{\left(1-\frac{1}{ k }\right)}$
$9( k -1)^3=4 k ( k -1)+1$
$k =2$
View full question & answer→MCQ 2281 Mark
Let $a _{ n }$ be the $n ^{\text {th }}$ term of the series $5+8+14+23$ $+35+50+\ldots$ and $S _{ n }=\sum \limits_{ k =1}^{ n } a _{ k }$. Then $S _{30}- a _{40}$ is equal to
- A
$11310$
- B
$11280$
- ✓
$11290$
- D
$11260$
AnswerCorrect option: C. $11290$
c
$S _{ n }=5+8+14+23+35+50+\ldots+a_n$
$S _{ n }=5+8+14+23+35+\ldots+ a _{ n }$
$O =5+3+6+9+12+15+\ldots . a _{ n }$
$a _{ n }=5+(3+6+9+\ldots( n -1) \text { terms })$
$a _{ n }=\frac{3 n ^2-3 n +10}{2}$
$a _{40}=\frac{3(40)^2-3(40)+10}{2}=2345$
$S _{30}=\frac{3 \sum \limits_{n=1}^{30} n ^2-3 \sum \limits_{ n =1}^{30} n +10 \sum \limits_{ n =1}^{30} 1}{2}$
$=\frac{\frac{3 \times 30 \times 31 \times 61}{6}-\frac{3 \times 30 \times 31}{2}+10 \times 30}{2}$
$S _{30}=13635$
$S _{30}- a _{40}=13635-2345$
$=11290$
View full question & answer→MCQ 2291 Mark
Let $S_{ k }=\frac{1+2+\ldots .+ K }{ K }$ and $\sum_{j=1}^n S_j^2=\frac{n}{A}\left( Bn ^2+ Cn + D \right)$, where $A , B , C , D \in N$ and $A$ has least value. Then
- ✓
$A + B$ is divisible by $D$
- B
$A+B=5(D-C)$
- C
$A + C + D$ is not divisible by $B$
- D
$A + B + C + D$ is divisible by $5$
AnswerCorrect option: A. $A + B$ is divisible by $D$
a
$S _{ k }=\frac{ k +1}{2}$
$S _{ k }^2=\frac{ k ^2+1+2 k }{4}$
$\therefore \sum \limits_{ j -1}^{ n } S _{ j }^2=\frac{1}{4}\left[\frac{ n ( n +1)(2 n +1)}{6}+ n + n ( n +1)\right]$
$=\frac{ n }{4}\left[\frac{( n +1)(2 n +1)}{6}+1+ n +1\right]$
$=\frac{ n }{4}\left[\frac{2 n ^2+3 n +1}{6}+ n +2\right]$
$=\frac{ n }{4}\left[\frac{2 n ^2+9 n +13}{6}\right]=\frac{ n }{24}\left[2 n ^2+9 n +13\right]$
$A =24, B =2, C =9, D =13$
View full question & answer→MCQ 2301 Mark
If $\operatorname{gcd}( m , n )=1$ and $1^2-2^2+3^2-4^2+\ldots \ldots$ $+(2021)^2-(2022)^2+(2023)^2=1012 m ^2 n$, then $m ^2- n ^2$ is equal to
Answerb
$1^2-2^2+3^2-4^2+\ldots(2021)^2-(2022)^2+(2023)^2=1012$
$m ^2 n$
$=(1-2)(1+2)+(3-4)(3+4)+\ldots+(2021-2022)$
$+2022)+(2023)^2$
$=(-1)(1+2+3+4+\ldots .+2022)+(2023)^2$
$=(-1) \cdot \frac{(2022)(2023)}{2}+(2023)^2$
$=2023(2023-1011)=2023 \times 1012$
$m ^2 n =2023=17^2 .7$
$m =17, n =7$
$m ^2- n ^2=17^2-7^2=240$
View full question & answer→MCQ 2311 Mark
The sum of the first $20$ terms of the series $5+11+$ $19+29+41+\ldots$ is $..........$.
- A
$3450$
- B
$3250$
- C
$3420$
- ✓
$3520$
AnswerCorrect option: D. $3520$
d
$S _{20}=5+11+19+29+\ldots \ldots$
Let $T _{ r }=a r^2+ br + c$
$T _1= a + b + c =5$
$T _2=4 a +2 b + c =11$
$T _3=9 a +3 b + c =19$
$a =1, b =3, c =1$
Hence $S _{20}=\sum_{ r =1}^{20} r ^2+3 \sum_{ r =1}^{20} r +\sum_{ r =1}^{20} 1=3520$
View full question & answer→MCQ 2321 Mark
The sum to $10$ terms of the series $\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+\ldots$. is:-
- A
$\frac{59}{111}$
- ✓
$\frac{55}{111}$
- C
$\frac{56}{111}$
- D
$\frac{58}{111}$
AnswerCorrect option: B. $\frac{55}{111}$
b
$T_r= \frac{\left( r ^2+ r +1\right)-\left( r ^2- r +1\right)}{2\left( r ^4+ r ^2+1\right)}$
$\Rightarrow T _r=\frac{1}{2}\left[\frac{1}{ r ^2- r +1}-\frac{1}{ r ^2+ r +1}\right]$
$T_1=\frac{1}{2}\left[\frac{1}{1}-\frac{1}{3}\right]$
$T_2=\frac{1}{2}\left[\frac{1}{3}-\frac{1}{7}\right]$
$\begin{array}{c}T_3=\frac{1}{2}\left[\frac{1}{7}-\frac{1}{13}\right] \\ \vdots \\ T _{10}=\frac{1}{2}\left[\frac{1}{91}-\frac{1}{111}\right] \\ \Rightarrow \sum \limits_{ r =1}^{10} T _{ r }=\frac{1}{2}\left[1-\frac{1}{111}\right]=\frac{55}{111}\end{array}$
View full question & answer→MCQ 2331 Mark
The sum $1^2-2.3^2+3.5^2-4.7^2+5.9^2-\ldots +15.29^2$ is $.......$.
- A
$6950$
- B
$6956$
- C
$6953$
- ✓
$6952$
AnswerCorrect option: D. $6952$
d
Separating odd placed and even placed terms we get
$S =\left(1 \cdot 1^2+3 \cdot 5^2+\ldots .15 \cdot(29)^2\right)-\left(2 \cdot 3^2+4.7^2\right.+\ldots .+14 \cdot(27)^2$
$S =\sum \limits_{ n =1}^8(2 n -1)(4 n -3)^2-\sum_{ n =1}^7(2 n )(4 n -1)^2$
Applying summation formula we get
$=29856-22904=6952$
View full question & answer→MCQ 2341 Mark
If $a_n=\frac{-2}{4 n^2-16 n+15}$, then $a_1+a_2+\ldots \ldots+a_{25}$ is equal to :
- A
$\frac{51}{144}$
- B
$\frac{49}{138}$
- ✓
$\frac{50}{141}$
- D
$\frac{52}{147}$
AnswerCorrect option: C. $\frac{50}{141}$
c
If $a_n=\frac{-2}{4 n^2-16 n+15}$ then $a_1+a_2+\ldots \ldots \ldots a_{25}$
$\sum \limits_{n=1}^{25} a_n=\sum \frac{-2}{4 n^2-16 n+15}$
$=\sum \frac{-2}{4 n^2-6 n-10 n+15}$
$=\sum \frac{-2}{2 n(2 n-3)-5(2 n-3)}$
$=\sum \frac{-2}{(2 n-3)(2 n-5)}$
$=\sum \frac{1}{2 n-3}-\frac{1}{2 n-5}$
$=\frac{1}{47}-\frac{1}{(-3)}$
$=\frac{50}{141}$
View full question & answer→MCQ 2351 Mark
If $\frac{1^3+2^3+3^3+\ldots \ldots \text {.upto } n \text { terms }}{1 \cdot 3+2 \cdot 5+3 \cdot 7+\ldots \ldots \text { upto } n \text { terms }}=\frac{9}{5}$, then the value of $n$ is
Answerd
$1^3+2^3+3^3 \ldots . .+n^3=\left(\frac{n(n+1)}{2}\right)^2$
$1 \cdot 3+2 \cdot 5+3 \cdot 7+\ldots \ldots+ n \text { terms }=$
$\sum \limits_{r=1}^n r(2 r+1)=\sum \limits_{r=1}^n\left(2 r^2+r\right)$
$=\frac{2 \cdot n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}$
$=\frac{ n ( n +1)}{6}(2(2 n+1)+3)$
$=\frac{ n ( n +1)}{2} \times \frac{(4 n +5)}{3}$
$=\frac{\frac{ n ^2( n +1)^2}{4}}{\frac{ n ( n +1)}{2} \times \frac{(4 n +5)}{3}}=\frac{9}{5}$
$\Rightarrow \frac{5 n(n+1)}{2}=\frac{9(4 n+5)}{3}$
$\Rightarrow 15 n ( n +1)=18(4 n +5)$
$\Rightarrow 15 n^2+15 n=72 n+90$
$\Rightarrow 15 n^2-57 n-90=0 \Rightarrow 5 n^2-19 n-30=0$
$\Rightarrow(n-5)(5 n+6)=0$
$\Rightarrow n=\frac{-6}{5} \text { or } 5$
$\Rightarrow n=5$
View full question & answer→MCQ 2361 Mark
Let $N$ be the sum of the numbers appeared when two fair dice are rolled and let the probability that $N -2, \sqrt{3 N }, N +2$ are in geometric progression be $\frac{ k }{48}$. Then the value of $k$ is
Answerb
$n ( s )=36$
Given : $N -2, \sqrt{3 N }, N +2$ are in G.P.
$3 N =( N -2)( N +2)$
$3 N = N ^2-4$
$\Rightarrow N ^2-3 N -4=0$
$( N -4)( N +1)=0 \Rightarrow N =4 \text { or } N =-1 \text { rejected }$
$( Sum =4) \equiv\{(1,3),(3,1),(2,2)\}$
$n ( A )=3$
$( A )=\frac{3}{36}=\frac{1}{12}=\frac{4}{48} \Rightarrow k =4$
View full question & answer→MCQ 2371 Mark
Let $f(x)$ be a function such that $f(x+y)=f(x) \cdot f(y)$ for all $x , y \in N$. If $f (1)=3$ and $\sum \limits_{ k =1}^{ n } f ( k )=3279$, then the value of $n$ is $.........$
Answerc
$f ( x + y )= f ( x ) \cdot f ( y ) \forall x , y \in N , f (1)=3$
$f (2)= f ^2(1)=3^2$
$f (3)= f (1) f (2)=3^3$
$f (4)=3^4$
$f ( k )=3^{ k }$
$\sum_{ k =1}^{ n } f ( k )=3279$
$f (1)+ f (2)+ f (3)+\ldots \ldots \ldots+ f ( k )=3279$
$3+3^2+3^3+\ldots \ldots \ldots 3^{ k }=3279$
$\frac{3\left(3^{ k }-1\right)}{3-1}=3279$
$\frac{3^{ k }-1}{2}=1093$
$3^{ k }-1=2186$
$3^{ k }=2187$
$k =7$
View full question & answer→MCQ 2381 Mark
If the sum of the series
$\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{2^2}-\frac{1}{2.3}+\frac{1}{3^2}\right)+$
$\left(\frac{1}{2^3}-\frac{1}{2^2 \cdot 3}+\frac{1}{2.3^2}-\frac{1}{3^3}\right)+$
$\left(\frac{1}{2^4}-\frac{1}{2^3 \cdot 3}+\frac{1}{2^2 \cdot 3^2}-\frac{1}{2 \cdot 3^3}+\frac{1}{3^4}\right)+\ldots$ is $\frac{\alpha}{\beta}$, where $\alpha$ and $\beta$ are co-prime, then $\alpha+3 \beta$ is equal to....
Answera
$P=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{2^2}-\frac{1}{2.3}+\frac{1}{3^2}\right)+$
$\left(\frac{1}{2^3}-\frac{1}{2^2 \cdot 3}+\frac{1}{2.3^2}-\frac{1}{3^3}\right)+\ldots \ldots$
$P\left(\frac{1}{2}+\frac{1}{3}\right)=\left(\frac{1}{2^2}-\frac{1}{3^2}\right)+\left(\frac{1}{2^3}+\frac{1}{3^3}\right)+\left(\frac{1}{2^4}-\frac{1}{3^4}\right)+\ldots$
$\frac{5 P}{6}=\frac{\frac{1}{4}}{1-\frac{1}{2}}-\frac{\frac{1}{9}}{1+\frac{1}{3}} \quad \therefore \alpha=1, \beta=2$
$\frac{5 P}{6}=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}$
$\therefore P=\frac{1}{2}=\frac{\alpha}{\beta}$
$\alpha+3 \beta=7$
View full question & answer→MCQ 2391 Mark
Let $\quad S=109+\frac{108}{5}+\frac{107}{5^2}+\ldots \ldots . .+\frac{2}{5^{107}}+\frac{1}{5^{108}}$. Then the value of $\left(16 S -(25)^{-34}\right)$ is equal to $............$.
- A
$2174$
- ✓
$2175$
- C
$2173$
- D
$2172$
AnswerCorrect option: B. $2175$
b
$S=109+\frac{108}{5}+\frac{107}{5^2} \ldots .+\frac{1}{5^{108}}$
$\frac{\frac{S}{5}=\frac{109}{5}+\frac{108}{5^2} \ldots \ldots+\frac{2}{5^{108}}+\frac{1}{5^{109}}}{\frac{4 S }{5}=109-\frac{1}{5}-\frac{1}{5^2} \ldots \ldots-\frac{1}{5^{108}}-\frac{1}{5^{109}}}$
$=109-\left(\frac{1}{5} \frac{\left(1-\frac{1}{5^{109}}\right)}{\left(1-\frac{1}{5}\right)}\right)$
$=109-\frac{1}{4}\left(1-\frac{1}{5^{109}}\right)$
$=109-\frac{1}{4}+\frac{1}{4} \times \frac{1}{5^{109}}$
$s =\frac{5}{4}\left(109-\frac{1}{4}+\frac{1}{4.5^{1099}}\right)$
$16 S =20 \times 109-5+\frac{1}{5^{108}}$
$16 S -(25)^{-54}=2180-5=2175$
View full question & answer→MCQ 2401 Mark
If $(20)^{19}+2(21)(20)^{18}+3(21)^2(20)^{17}+\ldots \ldots$. $+20(21)^{19}= k (20)^{19}$, then $k$ is equal to
Answerc
$\text { If }(20)^{19}+2(21)(20)^{18}+3(21)^2(20)^{17}+\ldots+20(21)^{19}=$
$k (20)^{19}$ then $k$ is
$20^{19}\left(1+2 \cdot\left(\frac{21}{20}\right)+3\left(\frac{21}{20}\right)^2+\ldots+20\left(\frac{21}{20}\right)^{19}\right)=k(20)^{19}$
$\Rightarrow k=1+2\left(\frac{21}{20}\right)+3\left(\frac{21}{20}\right)^2+\ldots .+20\left(\frac{21}{20}\right)^{19}$
$\Rightarrow k\left(\frac{21}{20}\right)=\frac{21}{20}+2 \cdot\left(\frac{21}{20}\right)^2+\ldots .$
$\ldots .+19\left(\frac{21}{20}\right)^{19}+20 \cdot\left(\frac{21}{20}\right)^{20}$
Subtracting equation $(2)$ from $(1)$
$\Rightarrow k\left(\frac{-1}{20}\right)=1+\frac{21}{20}+\left(\frac{21}{20}\right)^2+\ldots+\left(\frac{21}{20}\right)^{19}-20 \cdot\left(\frac{21}{20}\right)^{20}$
$\Rightarrow k\left(\frac{-1}{20}\right)=\frac{1\left(\left(\frac{21}{20}\right)^{20}-1\right)}{\left(\frac{21}{20}-1\right)}-20 \cdot\left(\frac{21}{20}\right)^{20}$
$\Rightarrow k\left(\frac{-1}{20}\right)=20\left(\frac{21}{20}\right)^{20}-20-20 \cdot\left(\frac{21}{20}\right)^{20}$
$\Rightarrow k\left(\frac{-1}{20}\right)=-20$
$\Rightarrow k=400$
View full question & answer→MCQ 2411 Mark
Let $[\alpha]$ denote the greatest integer $\leq \alpha$. Then $[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots .+[\sqrt{120}]$ is equal to.
Answerb
$[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots \ldots .[\sqrt{120}]$
$\Rightarrow 1+1+1+2+2+2+2+2+3+3+\ldots \ldots .+$
$3=7 \text { times }$
$+4+4+\ldots \ldots .+4=9 \text { times }+\ldots \ldots 10+10+$
$\ldots \ldots+10=21 \text { times }$
$\Rightarrow \sum_{r=1}^{10}(2 r+1) . r$
$\Rightarrow 2 \sum_{r=1}^{10} r^2+\sum_{r=1}^{10} r$
$\Rightarrow 2 \times \frac{10 \times 11 \times 21}{6}+\frac{10 \times 11}{2}$
$\Rightarrow 770+55$
$\Rightarrow 825$
View full question & answer→MCQ 2421 Mark
If $x=\sum \limits_{n=0}^{\infty} a^{n}, y=\sum\limits_{n=0}^{\infty} b^{n}, z=\sum\limits_{n=0}^{\infty} c^{n}$, where $a , b , c$ are in $A.P.$ and $|a| < 1,|b| < 1,|c| < 1$, $abc \neq 0$, then
- A
$x, y, z$ are in $A.P.$
- ✓
$\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in $A.P.$
- C
$x, y, z$ are in $G.P.$
- D
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1-(a+b+c)$
AnswerCorrect option: B. $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in $A.P.$
b
$x =1+ a + a ^{2}=\ldots \ldots \ldots .$
$x=\frac{1}{1-a} \Rightarrow a=1-\frac{1}{x}$
$y=\frac{1}{1-b} \Rightarrow b=1-\frac{1}{y}$
$z=\frac{1}{1-c} \Rightarrow c=1-\frac{1}{z}$
$a , b , c$ are in $A.P.$
$\Rightarrow 1-\frac{1}{x}, 1-\frac{1}{y}, 1-\frac{1}{z}$ are in $A.P.$
$\Rightarrow-\frac{1}{x},-\frac{1}{y},-\frac{1}{z}$ are in $A.P.$
$\Rightarrow \frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in $A.P.$
View full question & answer→MCQ 2431 Mark
If $a _{1}, a _{2}, a _{3} \ldots$ and $b _{1}, b _{2}, b _{3} \ldots$ are $A.P.$ and $a_{1}=2, a_{10}=3, a_{1} b_{1}=1=a_{10} b_{10}$ then $a_{4} b_{4}$ is equal to
- A
$\frac{35}{27}$
- B
$1$
- C
$\frac{27}{28}$
- ✓
$\frac{28}{27}$
AnswerCorrect option: D. $\frac{28}{27}$
d
$a_{1}, a_{2}, a_{3} \ldots \text { A.P. } ; a_{1}=2 ; a_{10}=3 ; d_{1}=\frac{1}{9}$
$b _{1}, b _{2}, b _{3}, \ldots$ $A.P.$ $; b _{1}=\frac{1}{2} ; b _{10}=\frac{1}{3} ; d _{2}=\frac{-1}{54}$
[Using $a_{1} b_{1}=1=a_{10} b_{10} ; d_{1}$ and $d_{2}$ are common differences respectively]
$a _{4} \cdot b _{4} =\left(2+3 d _{1}\right)\left(\frac{1}{2}+3 d _{2}\right)$
$=\left(2+\frac{1}{3}\right)\left(\frac{1}{2}-\frac{1}{18}\right)$
$=\left(\frac{7}{3}\right)\left(\frac{8}{18}\right)=\frac{28}{27}$
View full question & answer→MCQ 2441 Mark
If $\frac{1}{(20-a)(40-a)}+\frac{1}{(40-a)(60-a)}+\ldots \ldots+$ $\frac{1}{(180-a)(200-a)}=\frac{1}{256}$, then the maximum value of $a$ is.
Answerc
By splitting$\frac{1}{20}\left[\left(\frac{1}{20-a}-\frac{1}{40-a}\right)+\left(\frac{1}{40-a}-\frac{1}{60-a}\right)\right.$ $\left.+\ldots+\left(\frac{1}{180-a}-\frac{1}{200-a}\right)\right]$
$\frac{1}{20}\left(\frac{1}{20-a}-\frac{1}{200-a}\right)=\frac{1}{256}$
$(20-a)(200-a)=256 \times 9$
$a^{2}-220 a+1696=0$
$a=8,212$
Hence maximum value of a is $212 .$
View full question & answer→MCQ 2451 Mark
The $\operatorname{sum} \sum_{n=1}^{21} \frac{3}{(4 n-1)(4 n+3)}$ is equal to
- A
$\frac{7}{87}$
- ✓
$\frac{7}{29}$
- C
$\frac{14}{87}$
- D
$\frac{21}{29}$
AnswerCorrect option: B. $\frac{7}{29}$
b
$\sum_{n=1}^{21} \frac{3}{(4 n-1)(4 n+3)}=\frac{3}{4} \sum_{n=1}^{21} \frac{1}{4 n-1}-\frac{1}{4 n+3}$
$=\frac{3}{4}\left[\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{11}\right)+\ldots .+\left(\frac{1}{83}-\frac{1}{87}\right)\right]$
$=\frac{3}{4}\left[\frac{1}{3}-\frac{1}{87}\right]=\frac{3}{4} \frac{84}{3.87}=\frac{7}{29}$
View full question & answer→MCQ 2461 Mark
Let $3,6,9,12, \ldots$ upto $78$ terms and $5,9,13,17, \ldots$ upto $59$ terms be two series. Then, the sum of the terms common to both the series is equal to
- A
$2222$
- ✓
$2223$
- C
$2224$
- D
$2225$
AnswerCorrect option: B. $2223$
b
For series of common terms
$a =9, d =12, n =19$
$S _{19}=\frac{19}{2}[2(9)+18(12)]=2223$
View full question & answer→MCQ 2471 Mark
For a natural number $n$, let $a _{ n }=19^{ n }-12^{ n }$. Then, the value of $\frac{31 \alpha_{9}-\alpha_{10}}{57 \alpha_{8}}$ is
Answerd
$a _{ n }=19^{ n }-12^{ n }$
$\frac{31 \alpha_{9}-\alpha_{10}}{57 \alpha_{8}}=\frac{31\left(19^{9}-12^{9}\right)-\left(19^{10}-12^{10}\right)}{57 \alpha_{8}}$
$=\frac{19^{9}(31-19)-12^{9}(31-12)}{57 \alpha_{8}}$
$=\frac{19^{9} \cdot 12-12^{19} \cdot 19}{57 \alpha_{8}}$
$=\frac{12 \cdot 19\left(19^{8}-12^{8}\right)}{57 \alpha_{8}}$
$=4$
View full question & answer→MCQ 2481 Mark
Let $\alpha, \beta$ and $\gamma$ be three positive real numbers. Let $f ( x )=\alpha x ^{5}+\beta x ^{3}+\gamma x , x \in R \quad$ and $\quad g : R \rightarrow R$ be such that $g(f(x))=x$ for all $x \in R$. If $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ be in arithmetic progression with mean zero, then the value of $f\left(g\left(\frac{1}{n} \sum_{i=1}^{n} f\left(a_{i}\right)\right)\right)$ is equal to.
Answera
Consider a case when $\alpha=\beta=0$ then
$f(x)=y x$
$g(x)=\frac{x}{y}$
$\frac{1}{n} \sum_{i=1}^{n} f\left(a_{i}\right) \Rightarrow \frac{y}{n}\left(a_{1}+a_{2}+\ldots . .+a_{n}\right)$
$=0$
$f ( g (0)) \Rightarrow f (0)$
$0$
View full question & answer→MCQ 2491 Mark
Let the coefficients of the middle terms in the expansion of $\left(\frac{1}{\sqrt{6}}+\beta x\right)^{4},(1-3 \beta x)^{2}$ and $\left(1-\frac{\beta}{2} x\right)^{6}, \beta>0$, respectively form the first three terms of an $A.P.$ If $d$ is the common difference of this $A.P.$, then $50-\frac{2 d}{\beta^{2}}$ is equal to.
Answera
${ }^{4} C _{2} \times \frac{\beta^{2}}{6},-6 \beta,-{ }^{6} C _{3} \times \frac{\beta^{3}}{8}$ are in A.P
$\beta^{2}-\frac{5}{2} \beta^{3}=-12 \beta$
$\beta=\frac{12}{5} \text { or } \beta=-2 \therefore \beta=\frac{12}{5}$
$d =-\frac{72}{5}-\frac{144}{25}=-\frac{504}{25}$
$\therefore 50-\frac{2 d }{\beta^{2}}=57$
View full question & answer→MCQ 2501 Mark
Suppose $a_{1}, a_{2}, \ldots, a_{ n }, \ldots$ be an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms of the sum of first nine terms of the progression is $5: 17$ and $110< a_{15} < 120$ , then the sum of the first ten terms of the progression is equal to -
Answerb
$\frac{ S _{5}}{ S _{9}}=\frac{5}{17} \Rightarrow \frac{\frac{5}{2}(2 a+4 d)}{\frac{9}{2}(2 a+8 d)}=\frac{5}{17}$
$\Rightarrow d=4\,a$
$a_{15}=a+14 d=57\,a$
Now, $110< a _{15}<120$
$110<57\,a < 120$
$a =2 \therefore d =8$
$S _{10}=\frac{10}{2}(2 \times 2+9 \times 8)=380$
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