SEQUENCES AND SERIES - 1 (Arithmetic Progression)
Sequence
A sequence is a function of natural numbers with codomain as the set of real numbers. It is said to be finite or infinite according it has finite or infinite number of terms. Sequence $a _{1}, a _{2}, \ldots \ldots . . a _{n}$ is Series
usually denoted by $\left\{\mathrm{a} _{\mathrm{n}}\right\}$ or $<\mathrm{a} _{\mathrm{n}}>$
Series
By adding or subtracting the terms of a sequence we get a series.
Arithmetic Progression (A.P.)
A sequence is called an arithmetic progression, if the difference of two consecutive term is the same always.
i.e. $a _{n}-a _{n-1}=d$, (constant), $n _{\in} N$.
Here $\mathrm{d}$ is called the common difference (If $\mathrm{d}=0$ sequence is a constant sequence. if $\mathrm{d}>0$ the sequence is increasing; if $d<0$, the sequence is decreasing)
$\mathrm{n}^{\text {th }}$ term of an $\mathrm{A} . \mathrm{P}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}$
$\mathrm{n}^{\text {th }}$ term from end of an A.P $=\mathrm{a} _{\mathrm{n}}+(\mathrm{n}-1)(-\mathrm{d})$
where $a _{n}$ is the last term and $d$, the common difference of the A.P.
Sum of $n$ terms of an A.P
The sum $S _{n}$ of $n$ terms of an A.P is given by
$ S _{n}=\left\{\begin{array}{c} \frac{n}{2}[2 a+(n-1) d] \\ \frac{n}{2}\left(a _{1}+a _{n}\right) \end{array}\right. $
Where $a$ is the first term, $a _{n}$ last term, $d$ common difference
Note that a sequence is an A.P if and only if its $\mathrm{n}^{\text {th }}$ term is a linear expression in $\mathrm{n}$, and in that case its common difference is the coefficient of $n$.
Also sum to $\mathrm{n}$ terms is of the form $\mathrm{An}^{2}+\mathrm{Bn}$ where $\mathrm{A} \& \mathrm{~B}$ are constants, and the common difference of the A.P is $2 \mathrm{~A}$.
Note :
(i) d $=a _{n}-a _{n-1}$
$ =\left(S _{n}-S _{n-1}\right)-\left(S _{n-1}-S _{n-2}\right) $
$ =S _{n}-2 S _{n-1}+S _{n-2} $
(ii) $\mathrm{S} _{\mathrm{n}}-3 \mathrm{~S} _{\mathrm{n}-1}+3 \mathrm{~S} _{\mathrm{n}-2}-\mathrm{S} _{\mathrm{n}-3}$
$ \begin{aligned} & =\left(S _{n}-S _{n-1}^{n-1}\right)-2\left(S _{n-1}^{n-3}-S _{n-2}\right)+\left(S _{n-2}-S _{n-3}\right) \\ & =a _{n}-2 a _{n-1}+a _{n-2} \\ & =0 \text { as } a _{n-2,2} a _{n-1}, a _{n} \text { are in A.P } \end{aligned} $
Selection of terms in an A.P
In case of an odd number of terms the middle term is a and common difference $d$ while in case of even number of terms, middle terms are $\mathrm{a}-\mathrm{d}, \mathrm{a}+\mathrm{d}$ and common difference is $2 \mathrm{~d}$.
| No. of terms | Terms | Common difference. |
|---|---|---|
| 3 | $\mathrm{a}-\mathrm{d}, \mathrm{a}, \mathrm{a}+\mathrm{d}$ | d |
| 4 | $a-3 d, a-d, a+d, a+3 d$ | $2 d$ |
| 5 | $a-2d, a-d, a, a + d, a+2d$ | $d$ |
Insertion of Arithmeitc Means
If $a, b, c$ are in $A . P$, then $b=\frac{a+c}{2}$ is called the single arithmetic mean of $a \& c$. Let $a \& b$ be two given numbers and $A _{1}, A _{2}, \ldots \ldots \ldots . . . . A _{n}$ are $n$ A.M’s between them. Then $a, A _{1}, A _{2}, \ldots . A _{n}, b$ are in A.P. Common difference of this sequence $\mathrm{d}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}+1}$.
$A _{1}=a+d, A _{2}=a+2 d$ etc. we can find all the arithmetic means.
Properties of A.P.
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If $\mathrm{a}_1, \mathrm{a}_2, \mathrm{a}_3, \ldots \ldots .$. are in A.P; then $\mathrm{a}_1 \pm \mathrm{k}, \mathrm{a}_2 \pm \mathrm{k}, \mathrm{a}_3 \pm \mathrm{k}$, $\qquad$ are also in A.P.
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If $\mathrm{a}_1, \mathrm{a}_2, \mathrm{a}_3, \ldots \ldots \ldots \ldots \ldots . .$. are in A.P, then $\mathrm{a}_1 \lambda, \mathrm{a}_2 \lambda, \mathrm{a}_3 \lambda, \ldots \ldots \ldots \ldots \ldots \ldots . .$. and $\frac{\mathrm{a}_1}{\lambda}$, $\frac{\mathrm{a}_2}{\lambda}, \frac{\mathrm{a}_3}{\lambda} \ldots \ldots \ldots \ldots .$. are also in A.P $(\lambda \neq 0)$
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If ${a}_1,{a}_2,{a} _{{n}}$ are in A.P, then ${a} _{{n}},{a} _{{n}-1},.. {a}_2, {a}_1$ is also an A.P with common difference $(-{d})$
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If $\mathrm{a}_1, \mathrm{a}_2, \mathrm{a}_3, \ldots \ldots \ldots \ldots \ldots . . . \ldots$ and $\mathrm{b}_1, \mathrm{~b}_2, \mathrm{~b}_3, \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \ldots$ are two A.P.s then $a_1 \pm b_1, a_2 \pm b_2, a_3 \pm b_3, \ldots$. are also in A.P.
-
If $\mathrm{a}_1, \mathrm{a}_2, \mathrm{a}_3, \ldots \ldots \ldots \ldots \ldots$ and $\mathrm{b}_1, \mathrm{~b}_2, \mathrm{~b}_3, \ldots \ldots \ldots \ldots \ldots \ldots .$. are two A.P.s then $\mathrm{a}_1 \mathrm{~b}_1, \mathrm{a}_2 \mathrm{~b}_2, \mathrm{a}_3 \mathrm{~b}_3, \ldots \ldots \ldots \ldots .$. and $\frac{\mathrm{a}_1}{\mathrm{~b}_1}, \frac{\mathrm{a}_2}{\mathrm{~b}_2}, \frac{\mathrm{a}_3}{\mathrm{~b}_3}, \ldots \ldots \ldots \ldots \ldots . .$. are NOT in A.P.
-
If 3 numbers are in A.P we may take them as $\mathrm{a}-\mathrm{d}, \mathrm{a}, \mathrm{a}+\mathrm{d}$. If 4 numbers are in A.P, we can take them as $\mathrm{a}-3 \mathrm{~d}$, $\mathrm{a}-\mathrm{d}, \mathrm{a}+\mathrm{d}, \mathrm{a}+3 \mathrm{~d}$.
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In an arithmetic progression, sum of the terms equidistant form the beginning and end is a constant and equal to sum of first and last term.
$\quad$ $\quad$ Iie for $\left\{a _{n}\right\}, a _{1}+a _{n}=a _{2}+a _{n-1}=a _{3}+a _{n-2}=\ldots \ldots$
$\quad$ $\quad$ Also $\mathrm{a} _{\mathrm{r}}=\frac{\mathrm{a} _{\mathrm{r}-\mathrm{k}}+\mathrm{a} _{\mathrm{r}+\mathrm{k}}}{2}, 0 \leq \mathrm{k} \leq \mathrm{n}-\mathrm{r}$.
- Sum of $\mathrm{n}$ arithmetic means between two given numbers $\mathrm{a} \& \mathrm{~b}$ is $\mathrm{n}$ times the single A.M between them .
$\quad$ ie. $A _{1}+A _{2}+\ldots \ldots \ldots \ldots+A _{n}=n\left(\frac{a+b}{2}\right)$
- Also $S _{n}=a _{1}+a _{2}+\ldots \ldots+a _{n}=\left\{\begin{array}{l}n(\text { middle term); if } n \text { is odd. } \\ \frac{n}{2} \text { (sum of two middle terms); if } n \text { is even }\end{array}\right.$
Solved examples
1. If $x, y, z$ are real numbers satisfying the equation $25\left(9 x^{2}+y^{2}\right)+9 z^{2}-15(5 x y+y z+3 z x)=0$, then $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in
(a) A.P
(b) G.P
(c) H.P
(d) None of these
Show Answer
Solution:
We have $\left.(15 x)^{2}+(5 y)^{2}+(3 z)^{2}-(15 x)(5 y)-(5 y)(3 z)-(3 z) 15 x\right)=0$ or $(15 x-5 y)^{2}+(5 y-3 z)^{2}+(3 z-15 x)^{2}=0$
$ \because \mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}-\mathrm{ab}-\mathrm{bc}-\mathrm{ca}=\frac{1}{2}\left\{(\mathrm{a}-\mathrm{b})^{2}+(\mathrm{b}-\mathrm{c})^{2}+(\mathrm{c}-\mathrm{a})^{2}\right\} $
$\Rightarrow(15 x-5 y)=0,(5 y-3 z)=0,(3 z-15 x)=0$
$\Rightarrow 15 \mathrm{x}=5 \mathrm{y}=3 \mathrm{z} \Rightarrow \frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{3}=\frac{\mathrm{z}}{5}=\lambda$
$\therefore \mathrm{x}=\lambda, \mathrm{y}=3 \lambda, \mathrm{z}=5 \lambda$
So $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in A.P
Answer: (a)
2. The number of common terms of the two sequences 2,5,8,11 299 and $3,5,7,9,11 \ldots . .201$.
(a) 17
(b) 33
(c) 50
(d) 147
Show Answer
Solution:
Sequence of common terms is
$5,11,17 \ldots \ldots \ldots . .$. whose $n^{\text {th }}$ term is
$\mathrm{a} _{\mathrm{n}}=5+(\mathrm{n}-1) 6=6 \mathrm{n}-1$
$\mathrm{a} _{\mathrm{n}} \leq 201 \Rightarrow 6 \mathrm{n}-1 \leq 201$
$\Rightarrow \mathrm{n} \leq 33 \frac{2}{3} \therefore \mathrm{n}=33$
Answer: (b)
3. The value of $x+y+z=15$. If $a, x, y, z, b$ are in A.P, while the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is $\frac{5}{3}$. If $\frac{1}{a}$, $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}, \frac{1}{b}$ are in A.P, then
(a) $\mathrm{a}=1, \mathrm{~b}=9$
(b) $\mathrm{a}=9, \mathrm{~b}=1$
(c) can not find
(d) None of these
Show Answer
Solution:
$ \begin{aligned} & a+x+y+z+b=\frac{5}{2}(a+b) \\ & \Rightarrow x+y+z=\frac{3}{2}(a+b) \\ & \Rightarrow 15=\frac{3}{2}(a+b) \\ & \Rightarrow a+b=10 \end{aligned} $
Also $\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}+\frac{1}{\mathrm{~b}}=\frac{5}{2}\left(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}\right)$
$ \begin{gathered} \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{2}\left(\frac{1}{a}+\frac{1}{b}\right) \\ \frac{5}{3}=\frac{3}{2}\left(\frac{a+b}{a b}\right) \end{gathered} $
$\Rightarrow \frac{5}{3}=\frac{3}{2} \cdot \frac{10}{\mathrm{ab}} \Rightarrow \mathrm{ab}=9$
Solving we get $\mathrm{a}=1, \mathrm{~b}=9$ or $\mathrm{a}=9, \mathrm{~b}=1$
Answer:$=$ a, b
4. Let $\mathrm{a} _{1}, \mathrm{a} _{2}, \mathrm{a} _{3}$ $\ldots a _{11}$, be real numbers satisfying $a _{1}=15,27-2 a _{2}>0$ and $a _{k}=2 a _{k-1}-a _{k-2}$ for $\mathrm{k}=3,4, \ldots \ldots \ldots \ldots . .11$. If $\frac{\mathrm{a} _{1}{ }^{2}+\mathrm{a} _{2}{ }^{2}+\ldots \ldots .+\mathrm{a} _{11}{ }^{2}}{11}=90$, then the value of $\frac{\mathrm{a} _{1}+\mathrm{a} _{2}+\ldots \ldots .+\mathrm{a} _{11}}{11}$ is equal to
Show Answer
Solution:
$ \begin{aligned} & a _{1}=15, a _{k}=2 a _{k-1}-a _{k-2} \Rightarrow a _{1}, a _{2} \ldots \ldots \ldots a _{11} \text { are in A.P. } \\ & \therefore \frac{a _{1}^{2}+a _{2}^{2}+\ldots \ldots .+a _{11}^{2}}{11}=\frac{(15)^{2}+(15+d)^{2}+\ldots \ldots \ldots+(15+10 d)^{2}}{11}=90 \end{aligned} $
$\Rightarrow 7 \mathrm{~d}^{2}+30 \mathrm{~d}+27=0 \Rightarrow \mathrm{d}=-3, \frac{-9}{7}$
given $\mathrm{a} _{2}<\frac{27}{2} \therefore \mathrm{d}=-3 \& \mathrm{~d} \neq \frac{-9}{7}$
$ \left\{\begin{array}{l} \frac{\left(15^{2} \times 11\right)}{11}+\frac{d^{2}\left(1^{2}+-\cdot-+10^{2}\right)}{11}+\frac{3 o d(1+2+10)}{11} \\ \Rightarrow 225+350 d^{2}+150 d=90 \\ 35 d^{2}+150 d+135=0 \end{array}\right. $
$\Rightarrow \frac{\mathrm{a} _{1}+\mathrm{a} _{2}+\ldots \ldots+\mathrm{a} _{11}}{11}=\frac{11}{2} \frac{(30-3 \times 10)}{11}=0$
Answer: 0
5. Suppose A, B, C are defined as A $=a^{2} b+a b^{2}-a^{2} c-a c^{2}, B=b^{2} c+b c^{2}-a^{2} b-a b^{2}$ and $C=a^{2} c+c^{2} a-$ $\mathrm{cb}^{2}-\mathrm{c}^{2} \mathrm{~b}$, where $\mathrm{a}>\mathrm{b}>\mathrm{c}>0$ and the equation $\mathrm{Ax}^{2}+\mathrm{Bx}+\mathrm{C}=0$ has equal roots then $\frac{1}{\mathrm{a}}, \frac{1}{\mathrm{~b}}, \frac{1}{\mathrm{c}}$ are in
(a) A.P
(b) G.P
(c) H.P
(d) None of these
Show Answer
Solution:
$\because \mathrm{A}+\mathrm{B}+\mathrm{C}=0, \mathrm{x}=1$ is a root of $\mathrm{Ax}^{2}+\mathrm{Bx}+\mathrm{C}=0$
The other root $=1(\because$ roots are equal $)$
$\therefore 1 \times 1=\frac{\mathrm{C}}{\mathrm{A}} \Rightarrow \mathrm{C}=\mathrm{A}$
$a^{2} c+c^{2} a-c^{2}-c^{2} b=a^{2} b+a b^{2}-a^{2} c-a c^{2}$
$\mathrm{c}(\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b}+\mathrm{c})=\mathrm{a}(\mathrm{b}-\mathrm{c})(\mathrm{a}+\mathrm{b}+\mathrm{c})$
$\Rightarrow \mathrm{ac}-\mathrm{cb}=\mathrm{ab}-\mathrm{ac}$
$(\because \mathrm{a}+\mathrm{b}+\mathrm{c} \neq 0)$
$\Rightarrow 2 \mathrm{ac}=\mathrm{ab}+\mathrm{bc}$
$ \begin{aligned} & \Rightarrow \frac{2}{b}=\frac{1}{c}+\frac{1}{a} \\ & \Rightarrow \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \text { are in A.P. } \end{aligned} $
Answer: (a)
6. If a, b, c, d, are distinct integers in an increasing A.P such that $\mathrm{d}=\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}$, then $\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=$
(a) $-1$
(b) $0$
(c) $1$
(d) $2$
Show Answer
Solution:
$\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d} \in \mathrm{Z}$
Let $\mathrm{b}=\mathrm{a}+\lambda, \mathrm{c}=\mathrm{a}+2 \lambda, \mathrm{d}=\mathrm{a}+3 \lambda, \lambda \in \mathrm{Z}$
$a+3 \lambda=a^{2}+(a+\lambda)^{2}+(a+2 \lambda)^{2}$
$\Rightarrow 5 \lambda^{2}+3(2 \mathrm{a}-1) \lambda+3 \mathrm{a}^{2}-\mathrm{a}=0$
$\lambda$ is real
$\mathrm{D} \geq 0$
$9(2 \mathrm{a}-1)^{2}-4 \cdot 5 \cdot\left(3 \mathrm{a}^{2}-\mathrm{a}\right) \geq 0 \Rightarrow 24 \mathrm{a}^{2}+16 \mathrm{a}-9 \leq 0$
$\frac{-1}{3}-\frac{\sqrt{70}}{12} \leq \mathrm{a} \leq \frac{-1}{3}+\frac{\sqrt{70}}{12}$
$\because \mathrm{a} \in \mathrm{I} \quad \mathrm{a}=-1,0$
If $\mathrm{a}=-1, \quad \lambda=1$ and if $\mathrm{a}=0 \lambda \notin \mathrm{Z}$
$\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=\frac{4}{2}(\mathrm{a}+\mathrm{d})=2(2 \mathrm{a}+3 \lambda)=2(-2+3)=2$
Answer: (d)
7. Consider the sequence in the form of groups (1), (2,2), (3,3,3), (4,4,4,4), (5,5,5,5,5), the 2000th term of the sequence is not divisible by
(a) 3
(b) 9
(c) 7
(d) None of these
Show Answer
Solution:
Let us write the terms in the groups as follows:
$(1),(2,2),(3,3,3) \ldots \ldots .$. consisting of $1,2,3,4$, ..terms
Let 2000 th term fall in $\mathrm{n}^{\text {th }}$ group. Then
$1+2+3+$. $.+(n-1)<2000 \leq 1+2+$.
$\frac{\mathrm{n}(\mathrm{n}-1)}{2}<200 \leq \frac{\mathrm{n}(\mathrm{n}+1)}{2}$
$\Rightarrow \mathrm{n}(\mathrm{n}-1)<400 \leq \mathrm{n}(\mathrm{n}+1)$
$\mathrm{n}(\mathrm{n}-1)<400 \quad$ and $\mathrm{n}(\mathrm{n}+1) \geq 400$
$n^{2}-n-400<0 \quad$ and $n^{2}+n-400 \geq 0$
$\Rightarrow \mathrm{n}=63$
i.e. $2000^{\text {th }}$ term falls is $63 \mathrm{rd}$ group.
Also $2000^{\text {th }}$ term is 63
Answer: (d)
8. If $x^{18}=y^{21}=z^{28}$, then $3 \log _{y} \mathrm{x}, 3 \log _{z} \mathrm{y}, 7 \log _{\mathrm{x}} \mathrm{z}$ are in
(a) A.P
(b) G.P
(c) H.P
(d) None of these
Show Answer
Solution:
$ \begin{aligned} & \text { Let } x^{18}=y^{21}=z^{28}=\lambda \\ & 18 \log x=21 \log y=28 \log z=\log \lambda \end{aligned} $
$ \begin{aligned} & \log _{\mathrm{y}} \mathrm{x}=\frac{21}{18}, \log _{\mathrm{z}} \mathrm{y}=\frac{28}{21}, \log _{\mathrm{x}} \mathrm{z}=\frac{18}{28} \\ & \Rightarrow 3 \log _{\mathrm{y}} \mathrm{x}=\frac{7}{2}, 3 \log _{\mathrm{z}} \mathrm{y}=4,7 \log _{\mathrm{x}} \mathrm{z}=\frac{9}{2} \\ & \Rightarrow 3,3 \log _{\mathrm{y}} \mathrm{x}, 3 \log _{\mathrm{z}} \mathrm{y}, 7 \log _{\mathrm{x}} \mathrm{z} \text { are in A.P } \end{aligned} $
Answer: (a)
Practice questions
1 If $1, \log _{\mathrm{y}} \mathrm{x}, \log _{\mathrm{z}} \mathrm{y},-15 \log _{\mathrm{x}} \mathrm{z}$ are in A.P, then
(a) $\mathrm{z}^{3}=\mathrm{x}$
(b) $x=y^{-1}$
(c) $\mathrm{y}=\frac{1}{\mathrm{z}^{3}}$
(d) None of these
Show Answer
Answer: (a, b, c)2. If $5^{1+x}+5^{1-x}, \frac{a}{2}, 25^{x}+25^{-x}$ are three consecutive terms of an A.P, then a is
(a) $\leq 12$
(b) $\geq 12$
(c) $=12$
(d) None of these
Show Answer
Answer: (b)3. If $\sin \alpha, \sin ^{2} \alpha, 1, \sin ^{4} \alpha$ and $\sin ^{5} \alpha$ are in A.P, where $-\pi<\alpha<\pi$, then $\alpha$ lies in the interval
(a) $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$
(b) $\left(\frac{-\pi}{3}, \frac{\pi}{3}\right)$
(c) $\left(\frac{-\pi}{6}, \frac{\pi}{6}\right)$
(d) None of these
Show Answer
Answer: (d)4. If the roots of $x^{3}-12 x^{2}+39 x-28=0$ are in A.P, then their common difference will be
(a) $\pm 1$
(b) $\pm 2$
(c) $\pm 3$
(d) $\pm 4$
Show Answer
Answer: (c)5. If the sides of a right triangle are in A.P, then the sum of the sines of two acute angles is
(a) $\frac{7}{5}$
(b) $\frac{4}{\sqrt{3}}$
(c) $\sqrt{\frac{\sqrt{5}-1}{2}}+\sqrt{\frac{\sqrt{5}+1}{2}}$
(d) None of these
Show Answer
Answer: (a)6. Read the passage and answer the following questions. Two consecutive numbers from 1, 2, $3, \ldots \ldots \ldots, n$ are removed. The arithmetic mean of the remaining numbers is $\frac{105}{4}$.
(i). The value of $n$ lies in
(a) $[45,55]$
(b) $[52,60]$
(c) $[41,49]$
(d) None of these
Show Answer
Answer: (a)(ii). The removed numbers
(a) lie between 10 and 20
(b) are greater than 10
(c) are less than 15
(d) none of these
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Answer: (c)(iii). Sum of all numbers
(a) exceeds 1600
(b) is less than 1500
(c) lies between 1300 and 1500
(d) none of these
Show Answer
Answer: (b)
7. Concentric circles of radii $1,2,3, \ldots .100 \mathrm{~cm}$ 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 same color. The total area of the given regions in sq.cm is equal to
(a) $1000 \pi$
(b) $5050 \pi$
(c) $4950 \pi$
(d) $515 \pi$
Show Answer
Answer: (b)8. If three identical fair unbiased dice are thrown together such that the numbers $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$, where $a, b, c \in\{1,2,3,4,5,6\}$ appear on each of them respectively. If $\mathrm{r}$ represents all possible distinct cases, $\alpha$ represent the number of ways in which $a+b+c=9$ and $\beta$ represents the number of ways of obtaining $\mathrm{a}+\mathrm{b}+\mathrm{c}=8$, then match the following.
| Column I | Column II |
|---|---|
| (a) If $\alpha$ represents the common difference of an A.P such that the arithmetic mean of the squares of these quantities exceeds the square of A.M by 9 , then the number of terms of A.P are |
(p) ${ }^{5} \mathrm{C}_{2}$ |
| (b) $\sum _{r=m i n}^{\max (\alpha, \beta)}\left(\max (\alpha, \beta) \mathrm{C} _{\mathrm{r}}\right)-5$ | (q) 2 |
| (c) If 6th term in the expansion of $(\alpha+\beta)^{n}$ is the greatest term, then $\mathrm{n}$ is | (r) 1 |
| (d) $\left[\left(\frac{\gamma-\beta}{\gamma-\alpha}\right)\left(\frac{\alpha}{\beta}\right)\right]$ is equal to where [.] denotes the greatest integer function | (s) 10 |
Show Answer
Answer: a $\rarr$ q; b $\rarr$ q; c $\rarr$ p, s; d $\rarr$ r9. A person is to count 4500 currency notes. Let $\mathrm{a} _{\mathrm{n}}$ denotes the number of notes he counts in the $\mathrm{n}^{\text {th }}$ minute. If $\mathrm{a} _{1}=\mathrm{a} _{2}= ……….. a _{10}=150$ and $a _{10}, a _{11}………$ are in A.P with common difference 2 , then the time taken by him to count all notes is
(a) $24 \mathrm{~min}$
(b) $34 \mathrm{~min}$
(c) $125 \mathrm{~min}$
(d) $135 \mathrm{~min}$
Show Answer
Answer: (b)10. If $\mathrm{a} _{1}, \mathrm{a} _{2} \ldots \ldots \ldots \ldots . \mathrm{a} _{\mathrm{n}}$ are in A.P with common difference $\mathrm{d} \neq 0$ then $(\sin d) (\sec a { } _{1} \sec a _{2}+ \sec a _{2} \sec a _{3}+\ldots \ldots . . . \mathrm{\sec a} _{n-1} \sec a _{n} )$ is equal to
(a) $\operatorname{cota} _{\mathrm{n}}-\operatorname{cota} _{1}$
(b) $\operatorname{cota} _{1}-\operatorname{cota} _{n}$
(c) $\operatorname{tana} _{\mathrm{n}}-\operatorname{tana} _{1}$
(d) $\operatorname{tana} _{1}-\operatorname{tana} _{n}$
