datavorous/entrance-exam-dataset huggingface dataset:
The sum of lone pairs present on the central atom of the interhalogen IF5 and IF7 is ______

Options:

The correct answer is:
1

Solution:
In IF5, the central atom is iodine, which has 7 valence electrons. Iodine in this molecule has 5 bonding pairs and 1 lone pair. Therefore, the number of lone pairs on the central atom of IF5 is 1.

In IF7, the central atom is also iodine, which has 7 valence electrons. Iodine in this molecule has 7 bond pairs and 0 lone pairs. Therefore, the number of lone pairs on the central atom of IF7 is 0.

Therefore, the sum of the number of lone pairs in the central atom of IF5 and IF7 is 1+0=1..

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (10 Apr Shift 1)
The sum of $(n+1)$ terms of $\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots \ldots$ is

Options:
A $\frac{n}{n+1}$
B $\frac{2 n}{n+1}$
C $\frac{2}{n(n+1)}$
D $\frac{2(n+1)}{n+2}$

The correct answer is:
$\frac{2(n+1)}{n+2}$

Solution:
$T_n=\frac{1}{\left[\frac{n(n+1)}{2}\right]}=2\left[\frac{1}{n}-\frac{1}{n+1}\right]$
Put $n=1,2,3, \ldots,(n+1)$
$T_1=2\left[\frac{1}{1}-\frac{1}{2}\right], T_2=2\left[\frac{1}{2}-\frac{1}{3}\right]$\ldots\ldots\ldots
$T_{n+1}=2\left[\frac{1}{n+1}-\frac{1}{n+2}\right]$
Hence sum of $(n+1)$ terms $=\sum_{k=1}^{n+1} T_k$
$\Rightarrow S_{n+1}=2\left[1-\frac{1}{n+2}\right]$ $\Rightarrow S_{n+1}=\frac{2(n+1)}{(n+2)}$

Topic: Mathematics, Sequences and Series, JEE Main
The sum of $(n-1)$ terms of $1+(1+3)+(1+3+5)+\ldots \ldots$ is

Options:
A $\frac{n(n+1)(2 n+1)}{6}$
B $\frac{n^2(n+1)}{4}$
C $\frac{n(n-1)(2 n-1)}{6}$
D $n^2$

The correct answer is:
$\frac{n(n-1)(2 n-1)}{6}$

Solution:
Let $T_n$ be the $n^{\text {th }}$ term of the series $T_n=2 \sum n-\sum 1$
$\Rightarrow \quad T_n=\frac{2 n(n+1)}{2}-n=n^2$
$\therefore S_n=\sum_{k=1}^n\left(k^2\right)=\frac{n(n+1)(2 n+1)}{6}$
Hence sum of $(n-1)$ terms $S_{n-1}=\frac{(n-1) n(2 n-1)}{6}$

Topic: Mathematics, Sequences and Series, JEE Main
The sum of \( n \) terms of the series \( 5+7+13+31+85+\ldots \) is

Options:
A \( \frac{1}{2} n(2 n+1) \)
B \( \frac{1}{2}\left(3^{n}+8 n-1\right) \)
C \( \frac{1}{2}\left(3^{2 n}-2 n\right) \)
D \( \frac{1}{2}\left(3^{n}+4 n-1\right) \)

The correct answer is:
\( \frac{1}{2}\left(3^{n}+8 n-1\right) \)

Solution:
Let Sn=5+7+13+31+85+...+Tn...i

Where Tn is nth term.

This can be solved using method of difference as 7-5, 13-7, 31-13, 85-31...

2, 6, 18, 54 ... is a GP with common ratio 3.

Sn=5+7+13+31+85+...+Tn-1+Tn...ii Sn = 5+ 7+ 13+31+...+Tn-2+Tn-1+Tn...iii

Subtracting equation iii from ii, we get,

0=5+2+6+18+...(n-1) terms-Tn⇒Tn=5+23n-1-12

⇒Tn=3n-1+4

⇒Sn=∑n=1n3n-1+4

⇒Sn=3n-13-1+4n

⇒Sn=123n+8n-1

Topic: Mathematics, Sequences and Series, JEE Main
The sum of $n$ terms of the series whose $n^{\text {hth }}$ term is $n(n+1)$ is equal to

Options:
A $\frac{n(n+1)(n+2)}{3}$
B $\frac{(n+1)(n+2)(n+3)}{12}$
C $n^2(n+2)$
D $n(n+1)(n+2)$

The correct answer is:
$\frac{n(n+1)(n+2)}{3}$

Solution:
$T_n=n^2+n \Rightarrow S_n=\Sigma T_n=\Sigma n^2+\Sigma n$
$=\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}$

Topic: Mathematics, Sequences and Series, JEE Main
The sum of number of lone pairs of electrons present on the central atoms of XeO3,XeOF4 and XeF6 is

Options:

The correct answer is:
3

Solution:
Number of lone pairs on the central atom=Number of hybrid orbitals of the central atom-Number of sigma bonds of the central atom.

Number of hybrid orbitals or

Hybridisation for a molecule is given by = 12V+H-C+A

Where V= Number of valance electrons in central atom.

H = Number of surrounding monovalent atoms.

C= Cationic charge

A = Anionic charge.

XeF6:

Number of hybrid orbitals are 7.

Number of sigma bonds formed by the central atom are 6

Number of lone pairs on the central atom=7-6=1 lone pair

XeOF4:8+42=6

Number of hybrid orbitals are 6

Number of sigma bonds formed by the central atom are 5

Number of lone pairs on the central atom=6-5=1

XeO3:

Number of hybrid orbitals=4

Number of sigma bonds formed by the central atom are 3

Number of lone pairs on the central atom=4-3=1

Hence, the total number of lone pairs on the central atom of XeO3,XeOF4 and XeF6 are 3.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2022 (25 Jul Shift 2)
The sum of number of sigma and pi bonds formed between two carbon atoms in CaC2 are:

Options:

The correct answer is:
3

Solution:
CaC 2 → Ca 2+  +  C 2 2−
C 2 2− =   − C≡ C −

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The sum of oxidation state of the metals in Fe(CO)5, VO2+ and WO3 is

Options:

The correct answer is:
10

Solution:
In the complex Fe(CO)5, take the oxidation number of Fe be x . Carbonyl is a neutral ligand which doesn't have any charge. So the oxidation number of Fe would be

x-0×5 = 0= 0

Hence, the oxidation number of iron in this complex is 0.

let oxidation state of vanadium in VO2+ = x.

so, x + (-2) = +2 ⇒x = +2 + 2 = +4

Tungsten trioxide consists of one tungsten atom and three oxygen atoms. Tungsten is a d-block metal from group 6 and has an oxidation state +6 in the compound.

Then, Sum of oxidation state of metals in the given compounds =10

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The sum of oxidation states of two silver ions in AgNH32Ag(CN)2 complex is _____ .

Options:

The correct answer is:
2

Solution:
AgNH32Ag(CN)2→ AgNH32+ + Ag(CN)2- ( Ag+1) ( Ag+1)

Sum of oxidation states of two silver ions = 1 + 1 = 2

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (01 Sep Shift 2)
The sum of pπ−dπ bonds in the gases obtained by strong heating of ferrous sulphate is :

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The sum of possible values of x for tan-1x+1+cot-11x-1=tan-1831 is:

Options:
A -324
B -314
C -304
D -334

The correct answer is:
-324

Solution:
Given tan-1x+1+cot-11x-1=tan-1831

⇒tan-1x+1+tan-1x-1=tan-1831, ∴tan-1x=cot-11x, x>0

⇒tan-1x+1+x-11-x+1x-1=tan-1831

⇒tan-12x1-x+1x-1=tan-1831

Taking tangent both the sides we get,

(x+1)+(x-1)1-x2-1=831

⇒2x2-x2=831

⇒4x2+31x-8=0

⇒x=-8,14

But, if x=14

tan-1x+1∈0,π2

and cot-11x-1∈π2,π

⇒LHS>π2 and RHS<π2 (Not possible)

Hence, x=-8=-324.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The sum of protons, electrons and neutrons in the heaviest isotope of hydrogen is

Options:
A 6
B 5
C 4
D 3

The correct answer is:
4

Solution:
The heaviest isotope of hydrogen is Tritium. Tritium, also denoted as 'H or $\mathrm{T}$, contains:
1 proton
1 electron
2 neutrons 1
So, the sum of protons, electrons, and neutrons in Tritium is $1+1+2=4$. Therefore, the answer to your question is 4 .

Topic: Chemistry, Hydrogen, JEE Main
The sum of roots of the equation sin4x-cos2xsinx+2sin2x+sinx=0 for 0≤x≤3π is equal to

Options:
A 3π4
B 2π
C 4π
D 6π

The correct answer is:
6π

Solution:
sinxsin3x-cos2x+2sinx+1=0
sinxsin3x+2sinx+1-cos2x=0
sinxsin3x+sin2x+2sinx=0
sin2xsin2x+sinx+2=0
⇒sinx=0 as sin2x+sinx+2≠0
⇒x=nπ,n∈I
x∈0,3π,x=0,π,2π,3π
⇒ sum of roots =6π

Topic: Mathematics, Trigonometric Equations, JEE Main
The sum of series $\frac{1}{2 !}+\frac{1}{4 !}+\frac{1}{6 !}+\ldots$ is

Options:
A
$\frac{\left(\mathrm{e}^2-1\right)}{2}$

B
$\frac{(e-1)^2}{2 e}$

C
$\frac{\left(\mathrm{e}^2-1\right)}{2 \mathrm{e}}$

D
$\frac{\left(e^2-2\right)}{e}$

The correct answer is:
$\frac{(e-1)^2}{2 e}$

Solution:
$\frac{\mathrm{e}^\alpha+\mathrm{e}^{-\alpha}}{2}=1+\frac{\alpha^2}{2 !}+\frac{\alpha^4}{4 !}+\frac{\alpha^6}{6 !}+\ldots \ldots$
$\frac{\mathrm{e}^\alpha+\mathrm{e}^{-\alpha}}{2}-1=\frac{\alpha^2}{2 !}+\frac{\alpha^4}{4 !}+\frac{\alpha^6}{6 !}+\ldots \ldots$
put $\alpha=1$, we get
$\frac{(\mathrm{e}-1)^2}{2 \mathrm{e}}=\frac{1}{2 !}+\frac{1}{4 !}+\frac{1}{6 !}+\ldots \ldots \ldots \ldots$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2004
The sum of solutions of the equation cosx1+sinx=tan2x, x∈-π2,π2--π4,π4 is:

Options:
A π10
B -7π30
C -π15
D -11π30

The correct answer is:
-11π30

Solution:
We have,

cosx1+sinx=tan2x

where, x∈-π2,π2-π4,-π4

Case-1: When 0≤x<π4 & -π2<x<-π4.

cosx1+sinx=tan2x

⇒cosx1+sinx=sin2xcos2x

⇒cosx1+sinx=2sinxcosxcos2x-sin2x

⇒cosxcos2x-sin2x=2sinxcosx1+sinx

⇒cosx1-2sin2x=cosx2sinx+2sin2x

⇒cosx-4sin2x-2sinx+1=0

⇒cosx4sin2x+2sinx-1=0

Then,

cosx=0 (Not possible, since 0≤x<π4 & -π2<x<-π4)

And,

sinx=-2±258=-1±54

⇒x=π10,-3π10

Case-2: When π4<x<π2 and -π4<x<0, then

cosx1+sinx=-tan2x

⇒cosx1+sinx=-sin2xcos2x

⇒cosx1+sinx=-2sinxcosxcos2x-sin2x

⇒cosxcos2x-sin2x=-2sinxcosx1+sinx

⇒cosx1-2sin2x=cosx-2sinx-2sin2x

⇒cosx(1+2sinx)=0

⇒cosx=0⇒x∈ϕ

And, sinx=-12⇒x=-π6

Sum of solutions

=π10-3π10-π6=-11π30

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2021 (26 Aug Shift 1)
The sum of squares of all possible values of k, for which area of the region bounded by the parabolas 2y2=kx and ky2=2y−x is maximum, is equal to:

Options:

The correct answer is:
8

Solution:
Given: ky2=2y−x, 2y2=kx

Finding the point of intersection,

⇒ky2=2y−2y2k

⇒y=0 and ky=21−2yk

⇒ky+4yk=2

⇒y=2k+4k

⇒y=2kk2+4

So, the required area is given by,

A=∫02kk2+4y−ky22−2y2k⋅dy

⇒A=y22−k2+2k⋅y3302kk2+4

⇒A=2kk2+4212−k2+42k×13×2kk2+4

⇒A=16×4×1k+4k2

We know that, AM≥GM

⇒k+4k2≥2

⇒k+4k≥4

So, area is maximum when k=4k.

⇒k=2,−2

Hence, the sum of squares of all possible values of k is 8.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2024 (01 Feb Shift 2)
The sum of squares of deviation of a set is the least value when it is measured abouf

Options:
A Arithmetic mean
B Geometric mean
C Harmonic mean
D Median

The correct answer is:
Arithmetic mean

Solution:
The sum of squares of deviation of a set is the least value when it is measured about arithmetic mean.

Topic: Mathematics, Statistics, JEE Main
The sum of squares of deviations for $10$ observations taken from mean $50$ is $250$ . The co-efficient of variation is

Options:
A $50 \%$
B $10 \%$
C $40 \%$
D None of these

The correct answer is:
$10 \%$

Solution:
S.D. $(\sigma)=\sqrt{\frac{250}{10}}=\sqrt{25}=5$
Hence, coefficient of variation $=\frac{\sigma}{\text { mean }} \times 100$
$=\frac{5}{50} \times 100=10 \%$.

Topic: Mathematics, Statistics, JEE Main
The sum of squares of the abscissas of all the points on the line x+y=4 that lie at a unit distance from the line 4x+3y-10=0 is

Options:

The correct answer is:
58

Solution:
Any point on the line x+y=4 can be taken as x1,4-x1.

As it is at a unit distance from the line 4x+3y-10=0, we get,

4x1+34-x1-1042+32=1
⇒x1+2=5⇒x1+2=±5

⇒x1=3 or -7
∴ Required sum of the squares =9+49=58

Topic: Mathematics, Straight Lines, JEE Main
The sum of the 3rd and the 4th terms of a G.P. is 60 and the product of its first three terms is 1000. If the first term of this G.P. is positive, then its 7th term is:

Options:
A 320
B 640
C 2430
D 7290

The correct answer is:
320

Solution:
Let the 'G.P.' be
ar,a,ar,ar2
Given,

ar·a·ar=1000
⇒a3=1000
⇒a=10
Also, ar+ar2=60
⇒10r+r2=60
⇒r+r2=6
⇒r2+r-6=0
⇒r-2 r+3=0
So, r=2 or r=-3
As 1st term is positive ∴ r=2
∴ G.P. is '5', '10', '20'.......
Hence, 7th term is 5·26=320.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2015 (11 Apr Online)
The sum of the absolute maximum and absolute minimum values of the function fx=tan-1sinx-cosx in the interval 0,π is

Options:
A 0
B tan-112-π4
C cos-113-π4
D -π12

The correct answer is:
cos-113-π4

Solution:
Given,

fx=tan-1sinx-cosx

Now differentiating the function we get,

f'x=cosx+sinxsinx-cosx2+1

Now equating f'x=0 to find critical points,

cosx+sinxsinx-cosx2+1=0

⇒cosx+sinx=0

⇒tanx=-1

∴ x=3π4

Now checking the value of function at critical point and boundary point we get,

x	0	3π4	π

fx	-π4	tan-12	π4

So, we get,

fxmax=tan-12∴ fxmin=-π4

So, the sum will be =tan-12-π4

=cos-113-π4

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (28 Jul Shift 2)
The sum of the absolute minimum and the absolute maximum values of the function fx=3x-x2+2-x in the interval -1,2 is

Options:
A 17+32
B 17+52
C 5
D 9-172

The correct answer is:
17+32

Solution:
Given fx=x2-4x-2,∀x∈-1,3-172-x2+2x+2,∀x∈[3-172,2)

When x∈-1,3-172

f'x=2x-4=0⇒x=2

Now f2=2, f-1=3

f3-172=17-32

When x∈[3-172,2)

f'x=-2x+2

f'x=-2x-1

f'x=0 when x=1

f1=3

Hece, absolute minimum value =17-32

and absolute maximum value =3

Sum =17-32+3=17+32

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (26 Jun Shift 1)
The sum of the co-efficient of all even degree terms in x in the expansion of x+x3-16+x-x3-16, x>1 is equal to

Options:
A 26
B 32
C 24
D 29

The correct answer is:
24

Solution:
x+x3-16+x-x3-16

=26C0 x6 + 6C2x4x3-1+ 6C4 x2x3-12 + 6C6 x3-13

=26C0 x6 + 6C2 x7-x4+ 6C4 x2x6-2x3+1 +6C6 x9-3x6+3x3-1

∴ Sum of the coefficient of even power of x

=26C0-6C2+6C4+6C4-3-6C6

=21-15+15+15-3-1=24

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2019 (08 Apr Shift 1)
The sum of the co-efficient of all odd degree terms in the expansion of x+x3-15+x-x3-15, x>1 is

Options:
A 2
B -1
C 0
D 1

The correct answer is:
2

Solution:
x+x3-15+x-x3-15, (x>1)
fx=2x5+ 5C2x3x3-1+ 5C4xx3-12
Sum of coefficients of odd power of x
=f1-f-12
f1=2
f-1=2-1+ 5C2×2- 5C4×4
f-1=2-1+20-20=-2
⇒f1-f-12=2--22=2

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2018 (08 Apr)
The sum of the coefficient of $x^{2 / 3}$ and $x^{-2 / 5}$ in the binomial expansion of $\left(x^{2 / 3}+\frac{1}{2} x^{-2 / 5}\right)^9$ is

Options:
A $21 / 4$
B $63 / 16$
C $19 / 4$
D $69 / 16$

The correct answer is:
$21 / 4$

Solution:
$\begin{aligned} & \mathrm{T}_{\mathrm{r}+1}={ }^9 \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^{2 / 3}\right)^{9-\mathrm{r}}\left(\frac{\mathrm{x}^{-2 / 5}}{2}\right)^{\mathrm{r}} \\ & ={ }^9 \mathrm{C}_{\mathrm{r}}\left(\frac{1}{2}\right)^{\mathrm{r}}(\mathrm{r})^{\left(6 \frac{2 \mathrm{r}}{3} \frac{2 \mathrm{r}}{5}\right)}\end{aligned}$
for coefficient of $x^{2 / 3}$, put $6-\frac{2 r}{3}-\frac{2 r}{5}=\frac{2}{3}$
$\Rightarrow \mathrm{r}=5$
$\therefore$ Coefficient of $\mathrm{x}^{2 / 3}$ is $={ }^9 \mathrm{C}_5\left(\frac{1}{5}\right)^5$
For coefficient of $x^{-2 / 5}$, put $6-\frac{2 r}{3}-\frac{2 r}{5}=-\frac{2}{5}$
$\Rightarrow \mathrm{r}=6$
Coefficient of $\mathrm{x}^{-2 / 5}$ is ${ }^9 \mathrm{C}_6\left(\frac{1}{2}\right)^6$
$\text {Sum }={ }^9 \mathrm{C}_5\left(\frac{1}{2}\right)^5+{ }^9 \mathrm{C}_6\left(\frac{1}{2}\right)^6=\frac{21}{4}$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2024 (09 Apr Shift 2)
The sum, of the coefficients of the first 50 terms in the binomial expansion of 1-x100, is equal to

Options:
A 101C50
B 99C49
C -101C50
D -99C49

The correct answer is:
-99C49

Solution:
We know that 1-x100=C0100-C1100·x+C2100·x2-.......-C49100·x49+.......+C100100·x100

Now let, the sum of first fifty coefficients be S.

⇒S=C0100-C1100+C2100-.......-C49100

Let us substitute x=1 in the expansion.

⇒1-1100=C0100-C1100·1+C2100·12-.......-C49100·149+.......+C100100·1100

⇒0=C0100-C1100+C2100-.......-C49100+C50100-C51100+C52100.......+C100100

There are total of 101 terms and the middle term is T51=C50100.

As Crn=Cn-rn

⇒0=C0100-C1100+C2100-.......-C49100+C50100-C49100+C48100.......+C0100

⇒0=C50100+C0100-C1100+C2100-.......-C49100-C49100+C48100.......+C0100

⇒0=C50100+2C0100-C1100+C2100-.......-C49100

⇒-C501002=C0100-C1100+C2100-.......-C49100

⇒C0100-C1100+C2100-.......-C49100=-C501002

Therefore, the required answer is -C501002=-12×10050C4999=-C4999.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (12 Apr Shift 1)
The sum of the coefficients of three consecutive terms in the binomial expansion of 1+xn+2, which are in the ratio 1:3:5, is equal to

Options:
A 92
B 63
C 41
D 25

The correct answer is:
63

Solution:
Ratio of three consecutive terms in the binomial expansion of 1+xn+2 is

Cr-1n+2:Crn+2:Cr+1n+2::1:3:5

So,

⇒Crn+2Cr-1n+2=31

⇒n-r+3r=3

⇒n+3=4r ...1

And,

Cr+1n+2Crn+2=53

⇒n+2-rr+1=53

⇒3n+6-3r=5r+5

⇒3n+1=8r ...2

So, 34r-3+1=8r

⇒12r-9+1=8r

⇒4r=8

⇒r=2

So, n=5

Hence, we have

1+xn+2=1+x7

Sum of consecutive terms is

C17+C27+C37

=7+21+35=63

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (11 Apr Shift 2)
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 _____ .

Options:

The correct answer is:
321

Solution:
We have,

S1: 2,5,8,11,...,359

Common difference is d1=3

S2: 3,7,11,15...,399

Common difference is d2=4

S3:2,7,12,17...,197

Common difference is d3=5

Now taking LCM of common difference we get,

LCMd1, d2, d3=60

So, common terms are 47,107,167,

Hence, the sum will be 47+107+167=321.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (01 Feb Shift 2)
The sum of the digits in the unit's place of all the 4 - digit numbers formed by using the numbers 3, 4, 5 and 6, without repetition is :

Options:
A 18
B 36
C 108
D 432

The correct answer is:
108

Solution:
By fixing one digit at unit place we can form toal number without repetition is 3!

Sum of digits = 3! . 6 + 5 + 4 + 3

=618=108 ∵n!=n.n-1...3.2.1

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2014 (09 Apr Online)
The sum of the distinct real values of μ for which the vectors μi^+j^+k^, i^+μj^+k^, i^+j^+μk^ are co-planar, is

Options:
A 0
B -1
C 1
D 2

The correct answer is:
-1

Solution:
Given the three vectors are coplanar.

Hence, μ111μ111μ=0

⇒μμ2-1-1μ-1+11-μ=0

⇒μμ-1μ+1-2μ-1=0 ∵a2-b2=a-ba+b

⇒μ−1μ μ+1−2=0

⇒μ−1μ2+μ+2=0

⇒μ−1μ-1μ+2=0

⇒μ=1, 1, -2

Therefore, the sum of the distinct real values of μ=-2+1=-1.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2019 (12 Jan Shift 1)
The sum of the distinct real values of \( \mu \) for which the vectors \( \mu \hat{i}+\hat{j}+\hat{k}, \hat{i}+\mu \hat{j}+\hat{k}, \hat{i}+\hat{j}+\mu \hat{k} \) are co-planar, is

Options:
A \( 0 \)
B \( -1 \)
C \( 1 \)
D \( 2 \)

The correct answer is:
\( -1 \)

Solution:
Given the three vectors are coplanar.

Hence, μ111μ111μ=0

⇒μμ2-1-1μ-1+11-μ=0

⇒μμ-1μ+1-2μ-1=0 ∵a2-b2=a-ba+b

⇒μ−1μ μ+1−2=0

⇒μ−1μ2+μ+2=0

⇒μ−1μ-1μ+2=0

⇒μ=1, 1, -2

Therefore, the sum of the distinct real values of μ=-2+1=-1.

Topic: Mathematics, Vector Algebra, JEE Main
The sum of the divisors of 9600 is

Options:
A 3048
B 6120
C 31620
D 24384

The correct answer is:
31620

Solution:
9600=96Ã100=27Ã31Ã52
Number of divisors =8Ã2Ã3=48
Sum of divisors =20+21+22+.......+2730+3150+51+52
=28-12-1Ã32-13-1Ã53-15-1
=255Ã82Ã1244
=31620

Topic: Mathematics, Permutation Combination, JEE Main
The sum of the first 10 terms of the series 51⋅2⋅3+72⋅3⋅9+93⋅4⋅27+.... is

Options:
A 2+1310
B 1+111310
C 2-111310
D 1-111310

The correct answer is:
1-111310

Solution:
Tn=2n+3nn+13n
2n+3nn+1=An+Bn+1⇒A=3, B=-1
Tn=3n-1n+113n
=1n⋅3n-1-1n+13n=fn-fn+1 where, fn=1n⋅3n-1
⇒Sn=T1+T2+......+Tn
S10=T1+T2+......+T10=f1-f2+f2-f3+......+f10-f11
S10=f1-f11=11⋅30-111⋅310=1-111⋅310

Topic: Mathematics, Sequences and Series, JEE Main
The sum of the first 20 terms common between the series $3+7+11+15+$ and $1+6+11+$ $16+\ldots .$. is

Options:
A
4000

B
4020

C
4200

D
4220

The correct answer is:
4020

Solution:
Given $\mathrm{n}=20 ; \mathrm{S}_{20}=$ ?
Series (1) $\rightarrow 3,7, \underline{11}, 15,19,23,27, \underline{31}, 35$, $39,43,47$,
$\underline{51}, 55,59 \ldots$
Series (2) $\rightarrow 1,6, \underline{11}, 16,21,26, \underline{31}, 36,41$, $46, \underline{51}, 56$, $61,66,71$.
The common terms between both the series are $11,31,51,71 \ldots$
Above series forms an Arithmetic progression (A.P).
Therefore, first term (a) $=11$ and common difference $(\mathrm{d})=20$
Now, $\mathrm{S}_{\mathrm{n}}=\frac{n}{2}[2 a+(n-1) d]$
$$
\begin{aligned}
&\mathrm{S}_{20}=\frac{20}{2}[2 \times 11+(20-1) 20] \\
&\mathrm{S}_{20}=10[22+19 \times 20] \\
&\mathrm{S}_{20}=10 \times 402=4020 \\
&\therefore \mathrm{S}_{20}=4020
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2014 (11 Apr Online)
The sum of the first 20 terms common between the series 3+7+11+15+ ..... and 1+6+11+16+ ...... is

Options:
A  4000
B  4200
C  4220
D  4020

The correct answer is:
4020

Solution:
S 1 ≡3, 7, 11, 15,.......⇒ c.d=4⇒ d 1 =4
S 2 ≡1, 6, 11, 16,.......⇒ c.d=5⇒ d 2 =5
⇒Every 5th term of S1 and 4th term of S2 will be same
⇒ Terms common to both AP will have a=11 and d=20
Hence, S 20 = 20 2 [ ( 2×11 )+( 20−1 )20 ]
= 10×402
= 4020

Topic: Mathematics, Sequences and Series, JEE Main
The sum of the first 20 terms of the series 1+32+74+158+3116+… is

Options:
A 39+1219
B 38+1220
C 38+1219
D 39+1220

The correct answer is:
38+1219

Solution:
1+32+74+158+3116+…
=2-1+2-12+2-14+2-18+2-116+…upto 20 terms
=40-1+12+14+18+…upto 20 terms
=40-1-12201-12=40-2+1219=38+1219

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2018 (16 Apr Online)
The sum of the first $n$ terms of the series $1^2+2 \cdot 2^2+3^2+2 \cdot 4^2+5^2+2 \cdot 6^2+\ldots$ is $\frac{n(n+1)^2}{2}$ when $n$ is even. When $n$ is odd the sum is

Options:
A
$\frac{3 n(n+1)}{2}$

B
$\frac{\mathrm{n}^2(\mathrm{n}+1)}{2}$

C
$\frac{n(n+1)^2}{4}$

D
$\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right]^2$

The correct answer is:
$\frac{\mathrm{n}^2(\mathrm{n}+1)}{2}$

Solution:
If $n$ is odd then $(n-1)$ is even $\Rightarrow$ sum of odd terms $=\frac{(n-1) n^2}{2}+n^2=\frac{n^2(n+1)}{2}$.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2004
The sum of the first $n$ terms of the series $\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots \ldots \ldots$ is

Options:
A $2^n-n-1$
B $1-2^{-n}$
C $n+2^{-n}-1$
D $2^n-1$

The correct answer is:
$n+2^{-n}-1$

Solution:
The sum of the first $n$ terms is
$S_n=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{2^2}\right)+$ $\left(1-\frac{1}{2^3}\right)+\left(1-\frac{1}{2^4}\right)$ $+\ldots \ldots+\left(1-\frac{1}{2^n}\right)$
$=n-\left\{\frac{1}{2}+\frac{1}{2^2}+\ldots . .+\frac{1}{2^n}\right\}$
$=n-\frac{1}{2}\left(\frac{1-\frac{1}{2^n}}{1-\frac{1}{2}}\right)=$ $n-\left(1-\frac{1}{2^n}\right)=n-1+2^{-n}$
Trick: Check for $n=1,2$ i.e. $S_1=\frac{1}{2}, S_2=\frac{5}{4}$ and
$\Rightarrow S_1=\frac{1}{2}$ and
$S_2=2+2^{-2}-1=\frac{5}{4}$

Topic: Mathematics, Sequences and Series, JEE Main
The sum of the first $n$ terms of the series $\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots \ldots \ldots$. is

Options:
A $2^n-n-1$
B $1-2^{-n}$
C $n+2^{-n}-1$
D $2^n-1$

The correct answer is:
$n+2^{-n}-1$

Solution:
The sum of the first $n$ terms is
$S_n=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{2^3}\right)+\left(1-\frac{1}{2^4}\right)$ $+\ldots \ldots+\left(1-\frac{1}{2^n}\right)$
$=n-\left\{\frac{1}{2}+\frac{1}{2^2}+\ldots . .+\frac{1}{2^n}\right\}$
$=n-\frac{1}{2}\left(\frac{1-\frac{1}{2^n}}{1-\frac{1}{2}}\right)$$=n-\left(1-\frac{1}{2^n}\right)=n-1+2^{-n}$
Trick: Check for
$n=1,2$ i.e. $\quad S_1=\frac{1}{2}, S_2=\frac{5}{4}$
$\Rightarrow S_1=\frac{1}{2}$ and $S_2=2+2^{-2}-1=\frac{5}{4}$.

Topic: Mathematics, Sequences and Series, JEE Main
The sum of the first three terms of an arithmetic progression is 9 and the sum of their squares is 35. The sum of the first n terms of the series can be

Options:
A nn+1
B 2n2
C n4-n
D n6-n

The correct answer is:
n6-n

Solution:
Let a-d, a & a+d are three numbers in A.P.

Given, a-d+a+a+d=9⇒a=3
and a-d2+a2+a+d2=35
⇒3a2+2d2=35⇒d=±2
a=3, d=2⇒1,3,5,……..
Sn=n22+n-12=n2
a=3, d=-2⇒5,3,1,……..
Sn=n210+n-1-2=n5-n+1
=n6-n

Topic: Mathematics, Sequences and Series, JEE Main
The sum of the first three terms of G.P is S and their products is 27. Then all such S lie in

Options:
A (-∞,-9]∪[3,∞)
B [-3,∞)
C (-∞,-3]∪[9,∞)
D (-∞, 9]

The correct answer is:
(-∞,-3]∪[9,∞)

Solution:
Let terms are ar,a,ar

then a3=27 ⇒a=3

Now S=3r+3+3r

S=31r+r+3

r+1r≥2 ∀r>0 or r+1r≤-2 ∀r<0

S∈(-∞,-3]∪[9,∞)

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (02 Sep Shift 1)
The sum of the following series 1+6+912+22+327+1212+22+32+429+15(12+22+…+52)11+.... up to 15 terms, is:

Options:
A 7520
B 7510
C 7830
D 7820

The correct answer is:
7820

Solution:
tn=3n12+22+32+…+n22n+1=3n×nn+12n+16 2n+1
=12 n3+n2
∴S15=∑n=115tn=∑n=11512n3+n2

=12∑n=115n3+∑n=115n2

=12×15×1622+15×16×316 ∑n3=nn+122, ∑n2=nn+1n+26
=7200+620
=7820

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (09 Jan Shift 2)