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A sum of money under compound interest doubles itself in 4 years. In how many years will it become 16 times itself?

Option: 1
12 years

Option: 2
16 years

Option: 3
8 years

Option: 4
None of these

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Posted by

Kumar

A certain loan amounts, under compound interest, compounded annually earns an interest of Rs.1980 in the second year and Rs.2178 in the third year. How much interest did it earn in the first year?
Option: 1
Rs.1600

Option: 2
Rs.1800

Option: 3
Rs.1900

Option: 4
None of these

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Posted by

Ram

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If a, b and c are the greatest values of $^{19}C_{p},^{20}C_{q},^{21}C_{r}$ respectively, then:

Option: 1 $\frac{a}{11}=\frac{b}{22}=\frac{c}{42}$
Option: 2 $\frac{a}{10}=\frac{b}{11}=\frac{c}{42}$
Option: 3 $\frac{a}{11}=\frac{b}{22}=\frac{c}{21}$
Option: 4 $\frac{a}{10}=\frac{b}{11}=\frac{c}{21}$

Binomial Coefficient of the middle term is greatest.

Now,

$\\^nC_{r}\;\text{ is max at middle term}\\\begin{array}{l}{a=^{19} C_{p}=^{19} C_{10}=^{19} C_{9}} \\ {b=^{20} C_{q}=^{20} C_{10}} \\ {c=^{21} C_{r}=^{21} C_{10}=^{21} C_{11}}\end{array}$

$\frac{a}{^{19}C_9}=\frac{b}{ ^{20} \mathrm{C}_{10}}=\frac{c}{^{21} \mathrm{C}_{11}}$

$\frac{a}{^{19}C_9}=\frac{b}{\frac{20}{10} \cdot ^{19} \mathrm{C}_9}=\frac{c}{\frac{21}{11} \cdot \frac{20}{10} ^{19} \mathrm{C}_{9}}$

$\\\frac{a}{1}=\frac{b}{2}=\frac{c}{42 / 11}\\\frac{a}{11}=\frac{b}{22}=\frac{c}{42}$

Correct Option (1)

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Posted by

Kuldeep Maurya

If the sum of the coefficients of all even powers of x in the product $(1+x+x^{2}+....+x^{2n})(1-x+x^{2}-x^{3}+....+x^{2m})$ is 61, then n is equal to _________.
Option: 1 30
Option: 260
Option: 315
Option: 4 45

$\text { Let } (1+x+x^{2}+....+x^{2n})(1-x+x^{2}-x^{3}+....+x^{2n})=a_{0}+a_{1} x+a_{2} x^{2}+\ldots \ldots$

$\\Put \,\, x=1 \\ (2 n+1).1 =a_{0} + a_{1} + a_{2}+ \ldots \ldots \\ Put\, x =-1 \\ 1.(2 n+1)= a_{0} - a_{1} + a_{2}+\ldots \ldots \\ From (i)+(ii) \\ 4 n+2 =2 (a_{0}+a_{2}+\ldots ) \\ So,\,\, 2 n+1=61 \\ \Rightarrow n=30$

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Posted by

Kuldeep Maurya

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If $\alpha$ and $\beta$ be the coefficients of $x^{4}$ and $x^{2}$ respectively in the expansion of $\left ( x+\sqrt{x^{2}-1} \right )^{6}+\left ( x-\sqrt{x^{2}-1} \right )^{6},$ then :
Option: 1 $\alpha +\beta =30$
Option: 2 $\alpha -\beta =-132$
Option: 3 $\alpha +\beta =60$
Option: 4 $\alpha -\beta =60$

We know

$\mathrm{(x+y)^{n}+(x-y)^{n}=2\left[^{n} C_{0}\; x^{n}\; y^{0}+^{n} C_{2} \;x^{n-2}\; y^{2}+^{n} C_{4}\; x^{n-4} \;y^{4}+\ldots .\right]}$

Now,

$\left ( x+\sqrt{x^{2}-1} \right )^{6}+\left ( x-\sqrt{x^{2}-1} \right )^{6}$

$={2\left[^{6} \mathrm{C}_{0} \mathrm{x}^{6}+^{6} \mathrm{C}_{2} \mathrm{x}^{4}\left(\mathrm{x}^{2}-1\right)+^{6} \mathrm{C}_{4} \mathrm{x}^{2}\left(\mathrm{x}^{2}-1\right)^{2}+^{6} \mathrm{C}_{6}\left(\mathrm{x}^{2}-1\right)^{3}\right]} \\ {\quad=2\left[\mathrm{x}^{6}+15\left(\mathrm{x}^{6}-\mathrm{x}^{4}\right)+15 \mathrm{x}^{2}\left(\mathrm{x}^{4}-2 \mathrm{x}^{2}+1\right)+\left(\mathrm{x}^{6}-1-3 \mathrm{x}^{4}+3 \mathrm{x}^{2}\right)\right]} \\ {\quad=2\left(32 \mathrm{x}^{6}-48 \mathrm{x}^{4}+18 \mathrm{x}^{2}-1\right)} \\ {\quad\alpha=-96 \text { and } \beta=36} \\ {\therefore \quad \alpha-\beta=-132}$ Correct Option (2)

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Posted by

vishal kumar

The number of ordered pairs (r,k) for which $6\cdot ^{35}C_{r}=(k^{2}-3)\cdot ^{36}C_{r+1},$ where k is an integer, is :
Option: 1 $4$
Option: 2 6
Option: 3 2
Option: 4 3

As we have learnt

$^{n} C_{r}=\frac{n}{r} \cdot^{n-1} C_{r-1}$

Now,

$^{36}C_{r+1} (k^{2}-3)={^{35}C_{r}}\times {6}$

$\frac{36}{r+1}. ^{35}C_{r} (k^{2}-3)={^{35}C_{r}}\times {6}$

${k^{2}-3=\frac{r+1}{6} \Rightarrow k^{2}=3+\frac{r+1}{6}}$

r should be less than or equal to 35

Hence for k to be an integer, r can be 5 and 35

For r=5 we get k = -2, 2

For r=35 we get k=-3,3

We get 4 ordered pair (5,-2), (5,2), (35,-3), (35, 3)

Correct Option (1)

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Posted by

Ritika Jonwal

NEET 2024 Most scoring concepts

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The coefficient of $x^{7}$ in the expression $(1+x)^{10}+x(1+x)^{9}+x^{2}(1+x)^{8}+....+x^{10}$ is :
Option: 1 $420$
Option: 2 $330$
Option: 3 $210$
Option: 4 $120$

Binomial Theorem

$(x+y)^{n}=^{n} C_{0} x^{n}+^{n} C_{1} x^{n-1} y+^{n} C_{2} x^{n-2} y^{2}+\cdots+^{n} C_{n} y^{n}\;\;where,\;n\in \mathbb{N}$

Now,

Given series S is a GP, with a = (1+x)¹⁰ , r = x/(1+x), n = 11

So, S = $\frac{(1+x)^{10} \left ( \frac{x}{1+x}^{11}-1 \right )}{(\frac{x}{1+x})-1}$

= (1+x)¹¹- x¹¹

Hence coefficient of x⁷ is ¹¹C₇= 330

Correct option (2)

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Posted by

Ritika Jonwal

$\sum_{k=0}^{20}\left({ }^{20} \mathrm{C}_{\mathrm{k}}\right)^{2}$ is equal to :
Option: 1 ${ }^{40} \mathrm{C}_{21}$
Option: 2 ${ }^{41} \mathrm{C}_{20}$
Option: 3 ${ }^{40} C_{20}$
Option: 4 ${ }^{40} C_{19}$

$s= \sum_{k= 0}^{20}\left (\, ^{20}C_{k} \right )^{2}= \, ^{20}C_{0}\: ^{2}+ \, ^{20}C_{1}\: ^{2}+ \, ^{20}C_{2}\: ^{2}+\cdots ^{20}C_{20}\, ^{2}$
it is same as coefficient of $x^{20}$ in the expansion of $\left ( 1+x \right )^{20}\left ( x+1 \right )^{20}$
$= coeff\cdot of\, x^{20}\pi\left ( 1+x \right )^{40}$
$= \, ^{40}C_{20}$
option (3)