Sequences and Series Test

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Sequences and Series Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

In a triangle $$ABC$$,
$$acos^2(\frac{C}{2})+c\;cos^2(\frac{A}{2})=\dfrac{3b}{2}$$, then the sides $$a,b,c$$

A
Satisfy $$a+b=c$$

B
are in $$A.P.$$

C
are in $$G.P.$$

D
are in $$H.P.$$

Solution

$$a\cos^{2}\left(\dfrac{C}{2}\right)+C\cos^{2}\left(\dfrac{A}{2}\right)=\dfrac{3b}{2}$$

$$a\left[\dfrac{1+\cos C}{2}\right]+C\left[\dfrac{1+\cos A}{2}\right]=\dfrac{3b}{2}$$

$$\dfrac{a}{2}+\dfrac{c}{2}+\dfrac{1}{2}\left[a\cos C+C \cos A\right]=\dfrac{3b}{2}$$

$$a+c+a\cos C+ c \cos A=3b$$

$$2R\sin A+2R\sin C+2R\sin A\cos C+2R\sin C\cos A =3(2R)(\sin B)$$

$$\sin A+\sin C+\sin (A+C)=3\sin (B)$$

$$A+B+C=\pi\Rightarrow \sin(A+C)=\sin B$$

$$\sin A+\sin C+\sin B=3\sin (B)$$

$$\sin A+\sin C=2\sin B$$

$$2R\sin A+2R\sin C=2R\sin B$$

$$a+c=2b$$

$$\therefore a,b,c$$ are an $$A.P$$
Question 2
1 / -0

Let $$N=\{1,2,3,.......,10\},$$ then sum of product of number taken two at a time from the set $$N$$ is

A
$$55 \times 24$$

B
$$45\times 24$$

C
$$12 \times 45$$

D
$$55\times 12$$

Solution
Question 3
1 / -0

The sum $$\sum _{ i=0 }^{ m }{ \left( \begin{matrix} 10 \\ i \end{matrix} \right) \left( \begin{matrix} 20 \\ m-i \end{matrix} \right) }$$ (where $$\left( \begin{matrix} p \\ q \end{matrix} \right)=0$$ if $$p<q$$) is maximum where $$m$$ is

A
$$5$$

B
$$10$$

C
$$15$$

D
$$20$$

Solution
Question 4
1 / -0

$$1.3.4+2.5.8+3.6.9+$$ upto $$n$$ terms is equal to ________ .

A
$$\dfrac {n(n+1)(n+2)}{6}$$

B
$$\dfrac {n(n+1)(3n^{2}+23n+46)}{12}$$

C
$$\dfrac {n(27n^{3}+90n^{2}+45n-5)}{4}$$

D
$$\dfrac {n(n+1)(2n+1)}{6}$$

E
$$None\ of\ these$$

Solution
Question 5
1 / -0

If $$64a^2+36b^2=400, ab=4 $$ , then $$8a+6b$$ is:

A
26

B
28

C
22

D
None of the above

Solution

Given $$64a^2+36b^2=400,\quad ab=4$$
We know
$$(a+b)^2=a^2+b^2+2ab$$
therefore,
$$(8a+6b)^2=(8a)^2+(6b)^2+2(8a)(6a)$$
$$=64a^2+36b^2+96ab$$
$$=400+96\times4$$
$$=400+384$$
$$(8a+6b)^2=784$$
$$8a+6b=\sqrt{784}$$
$$8a+6b=28$$
Question 6
1 / -0

The sum of the series
$$1+2.2+3.2^{2}+4.2^{3}+5.2^{4}+...+1000.2^{999}$$ is

A
$$999.2^{999}-1$$

B
$$999.2^{1000}-1$$

C
$$999.2^{1000}+1$$

D
$$999.2^{999}+1$$

Solution

$$ S = 1+2.2+3.2^{2}+4.2^{3}+5.2^{4}+..... +1000.2^{999}$$ __ (1)
Multiply with '2' on both sides
$$ 2S = 1.2+2.2^{2}+3.2^{2}+4.2^{4}+5.2^{5}+... +999.2^{999}+1000.2^{1000}$$ __ (2)
subtracting equation (1) & (2)
$$ (S-2S) = [ 1+(2.2-1.2)+(3.2^{2}-2.2^{2})+(4.2^{3}-3.2^{3})+(5.2^{4}-4.2^{4})+$$ ...
$$+(1000.2^{999}-999.2^{999})-1000.2^{1000}$$
$$(S-2S) = 1+1.2+1.2^{2}+1.2^{3}+1.2^{4}+.... +1.2^{999}-1000.2^{1000}$$
$$-S = 1+(2+2^{2}+2^{3}+2^{4}+...+2^{999})-1000.2^{1000}$$
$$ \left [ \because a+ar+ar^{2}+...+ar^{n-1} = \dfrac{a.(r^{n}-1)}{r-1} \right ]$$
$$ a = 2, r = 2, n = 999 $$
$$ -S= \dfrac{1+2.(2^{999}-1)}{2-1} - 1000.2^{1000}$$
$$ -S = 1+2^{1000}-2-1000.2^{1000}$$
$$ -S = -1+2^{1000}(1-1000)$$
$$ -S = -1-999.2^{1000}$$
$$ S = 999.2^{1000}+1$$
Question 7
1 / -0

The sum of the series $$10.^{n}C_{0}+10^{2}.^{n}C_{1}+10^{3}.^{n}C_{2}+...10^{n+1}.^{n}C_{n}$$ is

A
$$11^{n}$$

B
$$10.11^{n}$$

C
$$11^{n+1}$$

D
$$11^{n}-1$$

Solution

$$ \rightarrow 10(^{n}C_{0}+10.^{n}C_{1}+...+10^{n}$$ $$^{n}C_{n})$$
$$ = 10\displaystyle(\sum_{r=1}^{n}$$ $$ ^{n}C_{r}(10)^{r}(1)^{n-r} )$$
$$ = 10(1+10)^{n} = 10.11^{n} = (B)$$
Question 8
1 / -0

If $$| x| < 1$$ , then the sum of series $$ 1+ 2x + 3x^2 + 4x^3 + .........\infty$$ will be

A
$$\dfrac{1}{1-x}$$

B
$$\dfrac{1}{1+x}$$

C
$$\dfrac{1}{(1+x)^2}$$

D
$$\dfrac{1}{(1-x)^2}$$

Solution

Let $$S=1+2x+3{x}^{2}+......\infty$$ $$\dots(1)$$

multiply by $$x$$ on both sides

$${x}{S}=x+2{x}^{2}+3{x}^{3}+.....\infty$$ $$\dots(2)$$

Subtract $$(1)$$ from $$(2)$$

$$S=1+2x+3{x}^{2}+......\infty$$

$${x}{S}=x+2{x}^{2}+3{x}^{3}+.....\infty$$
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$$S(1-x)=1+x+{x}^{2}+{x}^{3}+.....\infty$$

$$S(1-x)=\cfrac{1}{1-x}$$ [sum of infinte G.P $$S_{\infty}=\dfrac{1}{1-r}]$$

$$S=\cfrac{1}{(1-x)^{2}}$$
Question 9
1 / -0

There is a defined relationship between the pair of figure on either side of : :. Identify the relationship of the given pair and find be the missing figure.

A

B

C

D

Solution
Question 10
1 / -0

$$\displaystyle\sum^n_{r=1}\displaystyle\sum^{r-1}_{p=0}$$ $$^{n}C_r\cdot {^{r}C_p}\cdot 2^p$$ is equal to?

A
$$4^n-3^n+1$$

B
$$4^n-3^n-1$$

C
$$4^n-3^n+2$$

D
$$4^n-3^n$$

Solution