datavorous/entrance-exam-dataset huggingface dataset:
The sum of all real roots of equation e2x-46e2x-5ex+1=0 is

Options:
A ln4
B -ln3
C ln3
D ln5

The correct answer is:
-ln3

Solution:
Let ex=t so, e2x=t2

So, the given equation becomes, t2-46t2-5t+1=0

t2-4=0 (or) 6t2-5t+1=0

t2=4 (or) 6t2-3t-2t+1=0

t=±2 (or) 3t2 t-1-12 t-1=0

t=±2 (or) 3t-12t-1=0

t=±2 (or) t=13 (or) t=12

But, ex=t>0 ∵ ex>0;∀x∈R

∴ Only possible values for t are 2,13,12

When t=2, ex=2 ⇒x=loge2 (or) ln2

When t=13, ex=13 ⇒ x=loge13 or -ln3

When t=12, ex=12⇒x=loge12 or -ln2

∴ The sum of roots of given equation =ln2-ln3-ln2=-ln3.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2022 (24 Jun Shift 2)
The sum of all real values of x satisfying the equation x2-5x+5x2+4x-60=1 is

Options:
A 6
B 5
C 3
D -4

The correct answer is:
3

Solution:
The given equation will be true in 3 cases.
Case 1: when x2-5x+5=1
⇒x2-5x+4=0
⇒x=1, 4
Case 2: when x2+4x-60=0
⇒x=6, -10
Case 3: when x2-5x+5=-1 and x2+4x-60∈ even integers
Now, x2-5x+5=-1
⇒x=2, 3
Only x=2 satisfies the given condition,
Hence, sum of all real values of x is 1+4+6-10+2=3

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2016 (03 Apr)
The sum of all real values of x satisfying the equation x2-5x+5x2+4x-60=1 is

Options:
A 6
B 5
C 3
D -4

The correct answer is:
3

Solution:
The given equation will be true in 3 cases.
Case 1: when x2-5x+5=1
⇒x2-5x+4=0
⇒x=1, 4
Case 2: when x2+4x-60=0
⇒x=6, -10
Case 3: when x2-5x+5=-1 and x2+4x-60∈ even integers
Now, x2-5x+5=-1
⇒x=2, 3
Only x=2 satisfies the given condition,
Hence, sum of all real values of x is 1+4+6-10+2=3

Topic: Mathematics, Quadratic Equation, JEE Main
The sum of all the 4-digit distinct numbers that can be formed with the digits 1, 2, 2 and 3 is:

Options:
A 26664
B 122664
C 122234
D 22264

The correct answer is:
26664

Solution:
Digits are 1, 2, 2, 3

total distinct numbers 4!2!=12.

total numbers when

1 at unit place is 3.
2 at unit place is 6
3 at unit place is 3.

So, sum =(3+12+9)103+102+10+1

=(1111)×24

=26664

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The sum of all the elements in the set {n∈{1,2,……,100}∣ H.C.F. of n and 2040 is 1} is equal to __________.

Options:

The correct answer is:
1251

Solution:
The prime factorisation of the given number is 2040=23×3×5×17

The H.C.F. of n and 2040 is 1 if n is not a multiple of 2, 3, 5 and 17.

Hence n cannot be multiple of 2,3,5 and 17

Then sum is n(1)-(n(2)+n(3)+n(5)+n(17)-n(6)-n(10)-n(34)-n(15)-n(51)-n(85))+n(30))
Where n(a) means the sum of all numbers belonging to the set {1,2,…….100} which are divisible by a

=100×1012-2×50×512+3×33×342+5×20×212+17×5×62-6×16×172-10×10×112-34×2×32-15×6×72-51-85+180

=5050-2550-1683-1050-255+816+550+102+315+51+85-180

=1251

Alternate Solution:
Thus, the sum of all the required numbers taking the common elements only once, is 1+3+5+…..+99-3+9+15+21+……+99-5+25+35+55+65+85+95-17

Now, using the sum of n terms of an arithmetic progression i.e. Sn=n2a+l, where a is the first term and l is the last term.

=5021+99-1723+99-365-17

=502100-172102-365-17

=2500-867-365-17

=1251.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2021 (22 Jul Shift 1)
The sum of all the elements of the set α∈1,2,…..100:HCFα,24=1 is

Options:

The correct answer is:
1633

Solution:
Given S=1,2,3…100

Now finding the sum of S=100×1012

Prime factors of 24=23×3

Let nA= Multiples of 2

nB= Multiples of 3

nA∩B= Multiples of 2 & 3

So nA∪B=nA+nB−nA∩B

To have H.C.F to be 1 we need to subtract the sum of multiples of 2 & 3 from sum of set S to get required answer,

So required answer

=100×1012− Sum of nA∪B

=100×1012-2×50×512+332102−162×102 =1633

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2022 (24 Jun Shift 2)
The sum of all the local minimum values of the twice differentiable function f:R→R defined by fx=x3-3x2-3f"(2)2x+f"1 is:

Options:
A -22
B 5
C -27
D 0

The correct answer is:
-27

Solution:
fx=x3-3x2-3f"(2)2x+f"1 …i

⇒f'x=3x2-6x-32f"(2) …ii

⇒f"x=6x-6 …iii

Then,

f"(2)=12-6=6

f"(1)=0

Now,

f'x=3x2-6x-32f"(2)

⇒f'x=3x2-6x-32×6

⇒f'(x)=3x2-6x-9

For maxima/minima, we have

f'x=0

⇒3x2-6x-9=0

⇒x2-2x-3=0

⇒x2-3x+x-3=0

⇒x=-1 and 3

Now,

f"(x)=6x-6

f"(-1)=-12<0 (maxima)

f"(3)=12>0 (minima)

Again

fx=x3-3x2-3f"(2)2x+f"1

⇒fx=x3-3x2-32×6x+0

⇒f(x)=x3-3x2-9x

⇒f(3)=27-27-9×3=-27

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (20 Jul Shift 2)
The sum of all the real values of x Satisfying the equation 2x-1x2+5x-50=1 is:

Options:
A 16
B -5
C -4
D 14

The correct answer is:
-4

Solution:
Given 2x-1x2+5x-50=1

⇒x-1x2+5x-50=0, ∵20=1
⇒x-1x+10x-5=0
⇒x=1,5,-10
Sum of the real value of x=1+5-10=-4

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2017 (09 Apr Online)
The sum of all the roots of the equation x2-8x+15-2x+7=0 is

Options:
A 9-3
B 9+3
C 11-3
D 11+3

The correct answer is:
9+3

Solution:
Given,

x2-8x+15-2x+7=0

Now taking Case:I when,

x2-8x+15≥0⇒(x-3)(x-5)≥0

⇒x≤3 OR x≥5

x2-8x+15-2x+7=0

⇒x2-8x+15-2x+7=0

⇒x2-10x+22=0

⇒x=5+3, 5-3 rejected

So, α=5+3

Now taking Case:II when,

x2-8x+15-2x+7=0

⇒x2-8x+15<0⇒3<x<5

⇒x2-8x+15+2x-7=0

⇒x2-6x+8=0⇒x=4, 2 rejected

So, γ=4

Hence, α+γ=9+3

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2023 (06 Apr Shift 1)
The sum of all the solution of the equation $\cos \theta \cos \left(\frac{\pi}{3}+\theta\right) \cos \left(\frac{\pi}{3}-\theta\right)=\frac{1}{4}, \theta \in[0,6 \pi]$

Options:
A
$15 \pi$

B
$30 \pi$

C
$\frac{100 \pi}{3}$

D
None of these

The correct answer is:

$30 \pi$

Solution:
$$
\begin{aligned}
& 2 \cos \theta\left[\cos 120^{\circ}+\cos 2 \theta\right]=1 \\
\Rightarrow & 2 \cos \theta\left(-\frac{1}{2}+2 \cos ^2 \theta-1\right)=1 \\
\Rightarrow & 4 \cos ^3 \theta-3 \cos \theta-1=0 \\
\Rightarrow & \cos 3 \theta=1=\cos 0 \\
\Rightarrow & 3 \theta=2 n \pi \text { or } \theta=\frac{2 n \pi}{3}, n \in z
\end{aligned}
$$
Given the values so that $2 n$ does not exceed 18 .
Hence, the sum $=\frac{2 \pi}{3} \sum_1^9 n=\frac{2 \pi}{3} \times \frac{9(9+1)}{2}=30 \pi$

Topic: Mathematics, Trigonometric Equations, JEE Main
The sum of all the solutions in 0,100 for the equation sin⁡πx+cos⁡πx=0 is

Options:
A 2550
B 5025
C 2525
D 5050

The correct answer is:
5025

Solution:
sinâ¡Ïx+cosâ¡Ïx=0
tanâ¡Ïx=-1=tan3Ï4
Ïx=nÏ+3Ï4âx=n+34
Given, xâ0,100, hence n will range from 0 to 99.
x will be 34,74,114,.....,3994
Sn=n22a+n-1d
=5064+99Ã1=5025

Topic: Mathematics, Trigonometric Equations, JEE Main
The sum of all the solutions of the equation $(8)^{2 x}-16 \cdot(8)^x+48=0$ is :

Options:
A $1+\log _8(6)$
B $1+\log _6(8)$
C $\log _8(6)$
D $\log _8(4)$

The correct answer is:
$1+\log _8(6)$

Solution:
$\begin{aligned} & (8)^{2 x}-16 \cdot(8)^x+48=0 \\ & \text { Put } 8^{\mathrm{x}}=\mathrm{t} \\ & \mathrm{t}^2-16+48=0 \\ & \Rightarrow \mathrm{t}=4 \text { or } \mathrm{t}=12 \\ & \Rightarrow 8^x=4 \quad 8^x=12 \\ & \Rightarrow \mathrm{x}=\log _8 \mathrm{x} \quad \mathrm{x}=\log _8 12 \\ & \end{aligned}$
sum of solution $=\log _8 4+\log _8 12$
$\begin{aligned} & =\log _8 48=\log _8(6.8) \\ & =1+\log _8 6\end{aligned}$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2024 (08 Apr Shift 1)
The sum of all the values of λ for which the set x,y:x2+y2-6x+4y=12∩x,y:4x+3y=λ contains exactly one element is

Options:
A 31
B -31
C 12
D -19

The correct answer is:
12

Solution:
For exactly one element, the line should be a tangent to the circle
Given line L: 4x+3y=Î»
Circle S: x2+y2-6x+4y=12 has centre 3,-2, radius =5
Now, L will be a tangent to the circle if the distance of it from the centre of the circle is equal to the radius
â´p=r
43+3-2-Î»5=5âÎ»=31,-19
â Sum of all the values of Î» is equal to 12

Topic: Mathematics, Circle, JEE Main
The sum of all the values of p for which the lines x+y-1=0, px+4y+2=0 and 4x+py+7=0 are concurrent is equal to

Options:
A 0
B -9
C -13
D 4

The correct answer is:
-13

Solution:
11-1p424p7=0
⇒p2+9p-52=0
⇒p=-13, 4
p=4 is rejected as at p= 4 all 3 lines become parallel.

Topic: Mathematics, Straight Lines, JEE Main
The sum of all the values of Î» such that the lines x+2y-3=0, 3x-y-1=0 and 2Î»x+y-2=0 cannot form a triangle is

Options:

The correct answer is:
2.35

Solution:
Case–I: x+2y-3=0 & 2λx+y-2=0 are parallel
⇒2λ1=12⇒λ=0.25
Case–II: 3x-y-1=0 & 2λx+y-2 are parallel
⇒2λ3=1-1⇒λ=1.5
Case–III: Lines x+2y-3=0, 3x-y-1=0 & 2λx+y-2=0 are concurrent
⇒2λ1-212-33-1-1=0⇒-10λ-8+14=0⇒λ=0.6
⇒0.25+1.5+0.6=2.35

Topic: Mathematics, Straight Lines, JEE Main
The sum of all the values of x between 0 and 4Ï which satisfy the equation sinx8cos2x=1 is kÏ, then the value of k5 is equal to

Options:

The correct answer is:
2.4

Solution:
Given equation is sinxcos2x=122
⇒sin⁡xcos⁡x=122
Case (I): When x>0 i.e. x lies in 1st or 4th quadrant.
sin⁡xcos⁡x=122⇒sin⁡2x=12
⇒2x=π4,3π4,9π4,11π4,17π4,19π4
⇒x=π8,3π8,9π8,11π8,17π8,19π8
As x lies between 0 and 4π, cos⁡x>0⇒x=π8,3π8,17π8,19π8
Case (II): When cos⁡x<0 i.e., x lies in 2nd or 3rd quadrant.
sin⁡2x=-12⇒2x=5π4,7π4,13π4,15π4,21π4,23π4
⇒x=5π8,7π8,13π8,15π8,21π8,23π8
As x lies between 0 and 4π, cos⁡x<0⇒x=5π8,7π8,21π8,23π8
Sum of values of x=π81+3+17+19+5+7+21+23=96π8=12π
⇒k=12⇒k5=2.4

Topic: Mathematics, Trigonometric Equations, JEE Main
The sum of all those terms, of the arithmetic progression 3, 8, 13, ..., 373, which are not divisible by 3, is equal to ________.

Options:

The correct answer is:
9525

Solution:
Given sequence is 3,8,13,.....373.

Here a=3, d=5 and an=373

Now an=a+n-1d

⇒373=3+n-15

⇒n=75

We know that Sn=n2a+an

⇒S75=7523+373

⇒S75=14100

Now let us write the sequence of terms which are divisible by 3.

We get, 3,18,33,.....363

⇒363=3+n-115

⇒n=25

Now let us find the sum of terms divisible by 3.

⇒Sdiv by 3=2523+363=4575

Required sum =Sum of 75 terms-Sum of terms divisible by 3

=14100-4575

=9525.

Therefore, the required sum is 9525

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (10 Apr Shift 1)
The sum of all those terms which are rational numbers in the expansion of 213+31412 is:

Options:
A 89
B 27
C 35
D 43

The correct answer is:
43

Solution:
rth term in the expansion of 213+31412 is

Tr+1=Cr122r3·312-r4

Since, Tr+1 will be rational number, when r = 0,3,6,9,12 and r = 0,4,8,12

⇒r=0,12 for rational term of the expansion

Therefore, T1+T13=1×33+1×24×1

=27+16

=43.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (25 Jul Shift 2)
The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is

Options:
A 1356
B 1365
C 1256
D 1465

The correct answer is:
1356

Solution:
Two-digit numbers of the form 7λ+2 are 16, 23, 30,..., 93

These number are forming an arithmetic progression with first term a=16, common difference d=23-16=7 and last term l=93

And, we know that the last term of an arithmetic progression is given by l=a+n-1d

⇒93=16+n-17

⇒77=n-17

⇒n=12
Two-digit numbers of the form 7λ+5 are 12, 19, 26,..., 96

Again, using the last term of A.P., we get 96=12+n1-17

⇒84=n1-17

⇒n1=13

Now, the sum of n terms of an A.P. is Sn=n2a+l, we get the sum of all the above numbers as

=12216+93+13212+96

=654+702=1356.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (10 Jan Shift 1)
The sum of all values of θ∈0, π2 satisfying sin22θ+cos42θ=34 is

Options:
A π2
B 3π8
C 5π4
D π

The correct answer is:
π2

Solution:
Given sin22θ+cos42θ=34

Using sin2x+cos2x=1, we get

⇒1-cos22θ+cos42θ=34

Let cos22θ=t

⇒t2-t+14=0

⇒t-122=0

⇒t=12

⇒cos22θ=12

⇒2cos22θ-1=0

Using cos2A=2cos2A-1, we get

⇒cos4θ=0

If cosx=0, then x=2n+1π2, n∈Z.

⇒4θ=2n+1π2

⇒θ=2n+1π8

⇒θ=π8, 3π8 ∈ 0, π2

Hence, the sum of the values of θ is π8+3π8=π2.

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2019 (10 Jan Shift 1)
The sum of all values of α, for which the points whose position vectors are i^-2j^+3k^, 2i^-3j^+4k^, α+1i^+2k^ and 9i^+α-8j^+6k^ are coplanar, is equal to

Options:
A -2
B 2
C 6
D 4

The correct answer is:
2

Solution:
Let the given vectors be A→=i^-2j^+3k^, B→=2i^-3j^+4k^, C→=α+1i^+2k^ and D→=9i^+α-8j^+6k^

We know that if the vectors are coplanar then AB→AC→AD→=0

⇒AB→=2i^-3j^+4k^-i^-2j^+3k^=i^-j^+k^

⇒AC→=α+1i^+2k^-i^-2j^+3k^=αi^+2j^-k^

⇒AD→=9i^+α-8j^+6k^-i^-2j^+3k^=8i^+α-6j^+3k

Now,

⇒1-11α2-18α-63=0

⇒16+α-6+3α+8+α2-6α-16=0

On Simplifying we get,

⇒α2-2α-8=0

Therefore, sum of the roots is --2=2.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (06 Apr Shift 2)
The sum of all values of \( r \) in
\[
\left(\begin{array}{c}
18 \\
r-2
\end{array}\right)+2\left(\begin{array}{c}
18 \\
r-1
\end{array}\right)+\left(\begin{array}{c}
18 \\
r
\end{array}\right) \geq\left(\begin{array}{c}
20 \\
13
\end{array}\right) \text { must be }
\]

Options:

The correct answer is:
70

Solution:
18r-2+218r-1+18r≥2013 it means that

Cr-218+2Cr-118+Cr18≥C1320

⇒Cr-218+Cr-118+Cr-118+C18r≥C1320

⇒Cr-119+Cr19 ≥C1320

⇒Cr20≥C1320

⇒Cr20≥C720

Hence 7≤r≤13

∴ r=7, 8, 9, 10, 11, 12, 13

∴ Required sum

=7+8+9+10+11+12+13=70

Topic: Mathematics, Permutation Combination, JEE Main
The sum of all values of x in [0,2π], for which sinx+sin2x+sin3x+sin4x=0, is equal to :

Options:
A 8π
B 11π
C 12π
D 9π

The correct answer is:
9π

Solution:
Given,

sinx+sin2x+sin3x+sin4x=0

⇒(sinx+sin4x)+(sin2x+sin3x)=0

⇒2sin5x2cos3x2+2sin5x2cosx2=0

⇒2sin5x2cos3x2+cosx2=0

⇒2sin5x22cosxcosx2=0

2sin5x2=0⇒5x2=0,π,2π,3π,4π,5π

⇒x=0,2π5,4π5,6π5,8π5,2π

and cosx2=0⇒x2=π2⇒x=π

and cosx=0⇒x=π2,3π2

So sum =6π+π+2π=9π

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2021 (25 Jul Shift 1)
The sum of an infinite geometric progression is 54 times the sum of all the odd terms. If the first term is non-zero, then the common ratio of the geometric progression is

Options:
A 14
B 4
C 13
D 6

The correct answer is:
14

Solution:
a,ar,ar2,ar3,ar4,ar5,…,∞
S1=a+ar+ar2+ar3+…∞=a1-r
S2=a+ar2+ar4+ar6+…+∞=a1-r2
S1=54S2⇒a1-r=54a1-r2
⇒a1-r=54a1-r1+r
⇒1+r=54⇒r=14

Topic: Mathematics, Sequences and Series, JEE Main
The sum of an infinite geometric progression is \( 57 \) and the sum of their cubes is \( 9747 \), then what will be the common ratio of the geometric progression?

Options:
A \( \frac{1}{3} \)
B \( \frac{2}{3} \)
C \( \frac{1}{6} \)
D None of these

The correct answer is:
\( \frac{2}{3} \)

Solution:
Let a be the first term and r the common ratio of the G.P. Then

sum=57⇒a1-r=57 .......(i)

Sum of the cubes =9747

⇒a3+a3r3+a3r6+…=9747

⇒a31-r3=9747 .......(ii)

Dividing the cube of (i) by (ii), we

get a3(1-r)3·1-r3a3=(57)39747

⇒1-r3(1-r)3=19

⇒1+r+r2(1-r)2=19

⇒18r2-39r+18=0⇒(3r-2)(6r-9)=0⇒r=23 or r=32⇒r=23

(∵ r≠32 because -1<r<1 for an infinite G.P)

Topic: Mathematics, Sequences and Series, JEE Main
The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is $\frac{27}{19}$. Then the common ratio of this series is:

Options:
A $\frac{1}{3}$
B $\frac{2}{3}$
C $\frac{2}{9}$
D $\frac{4}{9}$

The correct answer is:
$\frac{2}{3}$

Solution:
Let the terms of infinite series are $a, a r, a r^{2}, a r^{3}, \ldots$

So, $\quad \frac{a}{1-r}=3$

Since, sum of cubes of its terms is $\frac{27}{19}$ that is sum of $a^{3}$,

$a^{3} r^{3}, \ldots \infty$ is $\frac{27}{19}$

So, $\frac{a^{3}}{1-r^{3}}=\frac{27}{19}$

$\Rightarrow \quad \frac{a}{1-r} \times \frac{a^{2}}{\left(1+r^{2}+r\right)}=\frac{27}{19}$

$\Rightarrow \frac{9\left(1+r^{2}-2 r\right) \times 3}{1+r^{2}+r}=\frac{27}{19}$

$\Rightarrow \quad 6 r^{2}-13 r+6=0$

$\Rightarrow(3 r-2)(2 r-3)=0$

$\Rightarrow r=\frac{2}{3},$ or $\frac{3}{2}$

As $|r| < 1$

So, $r=\frac{2}{3}$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (11 Jan Shift 1)
The sum of bridging carbonyls in WCO6 and Mn2CO10 is _____ .

Options:

The correct answer is:
0

Solution:
The structures of both WCO6 and Mn2CO10 are respectively:

Thus, from the structures, sum is zero.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (29 Jan Shift 1)
The sum of coefficients of integral powers of x in the binomial expansion of 1-2x50 is

Options:
A 1 2 2 50 + 1
B 1 2 3 50 + 1
C 1 2 3 50
D 1 2 3 50 - 1

The correct answer is:
1 2 3 50 + 1

Solution:
Integral Power Term is same as odd term in exapansion
Odd Term =1-2x50+1+2x502
Sum of coefficients=1-250+1+2502=1+3502

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2015 (04 Apr)
The sum of distinct values of λ for which the system of equations :

λ-1 x+3λ+1 y+2λz=0

λ-1 x+4λ-2 y+λ+3 z=0

2x+3λ+1 y+3 λ-1 z=0,

Has non-zero solutions, is ....... .

Options:

The correct answer is:
3

Solution:
λ-13λ+12λλ-14λ-2λ+323λ+13(λ-1)=0

R2→R2-R1

R3→R3-R1

λ-13λ+12λ0λ-3-λ+33-λ0λ-3=0

C1→C1+C3

3λ-13λ+12λ3-λλ-33-λ00λ-3=0

⇒λ-326λ=0 ⇒λ=0, 3

sum of values of λ=3

Topic: Mathematics, Determinants, JEE Main, JEE Main 2020 (06 Sep Shift 2)
The sum of first 9 terms of the series 131+13+231+3+13+23+331+3+5+... is

Options:
A 192
B 71
C 96
D 142

The correct answer is:
96

Solution:
Given, 131+13+231+3+13+23+331+3+5+...

∴ Tn=13+23+...+n31+3+...+2n-1

=n2n+124n2

=n+124
Sum of 9 terms

=∑n=19n+124
=1422+32+...+102
=1412+22+...+102-12

= 1 4 1 0 · 1 1 · 2 1 6 - 1

= 1 4 3 8 4 = 9 6

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2015 (04 Apr)
The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ......, is :

Options:
A 7 8 1 1 7 9 + 1 0 - 2 0
B 7 9 9 9 + 1 0 - 2 0
C 7 8 1 1 7 9 - 1 0 - 2 0
D 7 9 9 9 - 1 0 - 2 0

The correct answer is:
7 8 1 1 7 9 + 1 0 - 2 0

Solution:
Let, S=0·7+0·77+0·777+.....,

Here, number of terms is n.
⇒S=79(0·9+0·99+0·999+....)
⇒S=79910+99102+999103+...

⇒S=7910-110+102-1102+103-1103+...

⇒S=791+1+1+1+1+...+1-110+1102+1103+...+110n
⇒S=79n-110×1-110n1-110

For the sum of 20 terms, put n=20.
⇒S=7920-110×1-110201-110
⇒S=7920-1-10-209
⇒S=781179+10-20

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2013 (07 Apr)
The sum of first four terms of a geometric progression G.P. is 6512 and the sum of their respective reciprocals is 6518. If the product of first three terms of the G.P. is 1, and the third term is α, then 2α is _________.

Options:

The correct answer is:
3

Solution:
Let number are a,ar,ar2,ar3

ar4-1r-1=6512 …1

1a1r4-11r-1=6518

1ar31-r41-r=6518 …2

12⇒a2r3=32

and a3·r3=1

ar=1

ar2·r=32

r=32,a=23

So, third term=ar2=23×94

α=32

2α=3

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The sum of first four terms of a G.P. is 160 and the common ratio is 3 , then the $4^{\text {th }}$
term is

Options:
A 118
B 100
C 108
D 102

The correct answer is:
108

Solution:
$a, a r, a r^{2}, a r^{3}$
$2=3$
$\frac{a\left(\varepsilon^{4}-1\right)}{(\varepsilon-1)}=160$
Solving these we get $a=4$
$a \varepsilon^{3}=t_{4}=108$

Topic: Mathematics, Sequences and Series, JEE Main
The sum of first $n$ terms of the given series $1^2+2.2^2+3^2+2.4^2+5^2+2.6^2+$\ldots\ldots\ldots is $\frac{n(n+1)^2}{2}$, when $n$ is even. When $n$ is odd, the sum will be

Options:
A $\frac{n(n+1)^2}{2}$
B $\frac{1}{2} n^2(n+1)$
C $n(n+1)^2$
D None of these

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

Solution:
When $n$ is odd, the last term. i.e., the $n^{\text {th }}$ term will be $n^2$ in this case $n-1$ is even and so the sum of the first
$n-1$ terms of the series is obtained by replacing $n$ by $n-1$ in the given formula and so is $\frac{1}{2}(n-1) n^2$ Hence the sum of the $n$ terms $=$ (the sum of $n-1$ terms) + the $n^{\text {th }}$ term $=\frac{1}{2}(n-1) n^2+n^2=\frac{1}{2}(n+1) n^2$
Trick : Check for $n=1,3$. Here $S_1=1, S_3=18$ which gives (2).

Topic: Mathematics, Sequences and Series, JEE Main
The sum of four numbers in an arithmetic progression is 48 and the ratio of the product of the extremes to the product of the two middle terms is 27:35, then the largest term of this arithmetic progression is

Options:
A 10
B 12
C 14
D 18

The correct answer is:
18

Solution:
Let the A. P. be a-3d, a-d, a+d, a+3d
The sum of the terms =48=4a⇒a=12
Given, 12-3d12+3d12-d12+d=2735
⇒94-d4+d12-d12+d=2735
⇒16-d235=144-d23
⇒35d2-3d2=16×35-144×3
⇒32d2=1635-27=16×8
⇒d2=4⇒d=±2
So the numbers are: 6,10,14,18

Topic: Mathematics, Sequences and Series, JEE Main
The sum of infinite terms of the following series $1+\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}+\ldots \ldots .$. will be

Options:
A $\frac{3}{16}$
B $\frac{35}{8}$
C $\frac{35}{4}$
D $\frac{35}{16}$

The correct answer is:
$\frac{35}{16}$

Solution:
Let the sum to infinity of the arithmetico-geometric series be
$S=1+4 \cdot \frac{1}{5}+7 \cdot \frac{1}{5^2}+10 \cdot \frac{1}{5^3}+\ldots \ldots \ldots$
$\Rightarrow \quad \frac{1}{5} S=\frac{1}{5}+4 \cdot \frac{1}{5^2}+7 \cdot \frac{1}{5^3}+\ldots \ldots \ldots$.
Subtracting $\left(1-\frac{1}{5}\right) S=1+3 \cdot \frac{1}{5}+3 \cdot \frac{1}{5^2}+3 \cdot \frac{1}{5^3}+\ldots \ldots \ldots$
$=1+3\left(\frac{1}{5}+\frac{1}{5^2}+\ldots \ldots\right)$
$\Rightarrow \quad \frac{4}{5} \cdot S=1+3 \cdot \frac{1}{5}\left(\frac{1}{1-\frac{1}{5}}\right)$ $=1+\frac{3}{4}=\frac{7}{4} \Rightarrow S=\frac{35}{16}$
Aliter: Use $S=\left[1+\frac{r}{1-r} \times\right.$ diff. of A.P. $] \frac{1}{1-r}$

Topic: Mathematics, Sequences and Series, JEE Main
The sum of infinite terms of the following series $1+\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}+\ldots \ldots \ldots$. will be

Options:
A $\frac{3}{16}$
B $\frac{35}{8}$
C $\frac{35}{4}$
D $\frac{35}{16}$

The correct answer is:
$\frac{35}{16}$

Solution:
Let the sum to infinity of the arithmetico-geometric series be $S=1+4 \cdot \frac{1}{5}+7 \cdot \frac{1}{5^2}+10 \cdot \frac{1}{5^3}+\ldots \ldots \ldots$
$\Rightarrow \quad \frac{1}{5} S=\frac{1}{5}+4 \cdot \frac{1}{5^2}+7 \cdot \frac{1}{5^3}+\ldots \ldots \ldots$
Subtracting $\left(1-\frac{1}{5}\right) S=1+3 \cdot \frac{1}{5}+3 \cdot \frac{1}{5^2}+3 \cdot \frac{1}{5^3}+\ldots \ldots \cdot$
$=1+3\left(\frac{1}{5}+\frac{1}{5^2}+\ldots \ldots\right)$
$\Rightarrow \quad \frac{4}{5} \cdot S=1+3 \cdot \frac{1}{5}\left(\frac{1}{1-\frac{1}{5}}\right)$ $=1+\frac{3}{4}=\frac{7}{4} \Rightarrow S=\frac{35}{16}$
Aliter : Use direct formula
$S_{\infty}=\frac{a b}{1-r}+\frac{d b r}{(1-r)^2}$
Here $a=1, b=1, d=3, r=\frac{1}{5}$, therefore
$S_{\infty}=\frac{1}{1-\frac{1}{5}}+\frac{3 \times 1 \times \frac{1}{5}}{\left(1-\frac{1}{5}\right)^2}=\frac{5}{4}+\frac{\frac{3}{5}}{\frac{16}{25}}$ $=\frac{5}{4}+\frac{15}{16}=\frac{35}{16}$
Aliter: Use
$S=\left[1+\frac{r}{1-r} \times\right.$ diff. of A.P. $] \frac{1}{1-r}$

Topic: Mathematics, Sequences and Series, JEE Main
The sum of infinite terms of the sequence whose rth term is given by tr=1r+1r+3 is equal to

Options:
A 1
B 2
C 34
D 512

The correct answer is:
512

Solution:
tr=12r+3-r+1r+1r+3
=121r+1-1r+3
∑r=1∞tr=12∑r=1∞1r+1-1r+3
=1212-14+13-15+14-16+.…….
=1212+13=512.

Topic: Mathematics, Sequences and Series, JEE Main
The sum of infinitely many terms of a geometric progression is 3 times the sum of even terms. If the first term is non-zero, then the common ratio of the geometric progression is

Options:
A 12
B 2
C 32
D 52

The correct answer is:
12

Solution:
S1=a+ar+ar2+.……+arn-1+.…….∞
S2=ar+ar3+ar5+.……∞
S1=a1-r,S2=ar1-r2
Given S1=3S2
⇒a1-r=3ar1-r2
⇒1=3r1+r
⇒1+r=3r⇒r=12

Topic: Mathematics, Sequences and Series, JEE Main
The sum of infinity of a geometric progression is $\frac{4}{3}$ and the first term is $\frac{3}{4}$. The common ratio is

Options:
A $7 / 16$
B $9 / 16$
C $1 / 9$
D $7 / 9$

The correct answer is:
$7 / 16$

Solution:
According to condition
$\frac{3 / 4}{1-r}=\frac{4}{3} \Rightarrow r=\frac{7}{16}$

Topic: Mathematics, Sequences and Series, JEE Main
The sum of integers from 1 to 100 that are divisible by 2 or 5 is

Options:
A
3000

B
3050

C
3600

D
3250

The correct answer is:
3050

Solution:
Required sum $=(2+4+6+\ldots \ldots+100)+(5+10+15+\ldots \ldots+100)-(10+20+\ldots . .+100)$
$=2550+1050-530=3050$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2002