datavorous/entrance-exam-dataset huggingface dataset:
Let a1=8,a2,a3,….an be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170 , then the product of its middle two terms is _____ .

Options:

The correct answer is:
754

Solution:
Let the common difference of an A.P. be d, then

a1+a2+a3+a4=50

⇒4a+6d=50

⇒32+6d=50

⇒d=3

Also,

an-3+an-2+an-1+an=170

⇒4a+4n-10·d=170

⇒32+4n-10·3=170

⇒n=14

So, middle terms are

a7=a+6d=26, a8=a+7d=29

⇒a7·a8=754

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Let A=110310-310110 and B=1-i01, where i=-1. If M=ATBA, then the inverse of the matrix AM2023 AT is

Options:
A 1-2023i01
B 10-2023i1
C 102023i1
D 12023i01

The correct answer is:
12023i01

Solution:
Given,

A=110310-310110, B=1-i01

AAT=110310-310110110-310310110=1001

B2=1-i011-i01=1-2i01

B3=1-2i011-i01=1-3i01

So, we can write

B2023=1-2023i01

M=ATBA (given)

M=AATB=10011-i01=1-i01

M2=M·M=ATBAATBA=AAT.AAT.B2

=AATB2

M3=AATB3

M2023=AATB2023=B2023 (∵ AAT is an identity matrix)

∴ M2023=B2023=1-2023i01

Inverse of AM2023 AT is 12023i01

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Let a1, a2, a3,..........., a11 be real numbers satisfying a1=15, 27-2a2>0 and ak=2ak-1-ak-2 â k=3, 4, .......,11. If a 1 2 + a 2 2 + .... + a 1 1 2 1 1 = 9 0 , then the value of a 1 + a 2 + .... + a 1 1 1 1 is equal to

Options:

The correct answer is:
0

Solution:
a k = 2 a k - 1 - a k - 2
âa1, a2,...., a11 are in AP with let common difference be d
â´ a 1 2 + a 2 2 + â¦ + a 1 1 2 1 1 = 1 1 a 2 + 3 5 Ã 1 1 d 2 + 11 0 a d 1 1 = 9 0
â 2 2 5 + 3 5 d 2 + 1 5 0 d = 9 0
3 5 d 2 + 1 5 0 d + 1 3 5 = 0
â d = - 3 , - 9 7
Given, a 2 < 2 7 2 â´ d = - 3 and d â - 9 7

Hence, a1+a2+.......a11=1122Ã15+10(-3)=0

Topic: Mathematics, Sequences and Series, JEE Main
Let a1, a2, a3…,a10 be in G.P. with ai>0 for i=1, 2, …, 10 and S be the set of pairs r, k, r, k∈N (the set of natural numbers) for which
loge⁡a1r a2kloge⁡a2ra3kloge⁡a3ra4kloge⁡a4r a5kloge⁡a5ra6kloge⁡a6ra7kloge⁡a7ra8kloge⁡a8ra9kloge⁡a9ra10k=0
Then the number of elements in S, is:

Options:
A Infinitely many
B 4
C 10
D 2

The correct answer is:
Infinitely many

Solution:
C3→C3-C2,  C2→C2-C1 ( Let α is common ratio of GP)
logea1ra2klogeαr+klogeαr+klogea4ra5klogeαr+kloge⁡αr+klogea7ra8kloge⁡αr+kloge⁡αr+k=0 which is always true because C2 & C3 are identical.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (10 Jan Shift 2)
Let $a_1, a_2, a_3, \ldots, a_{11}$ be real numbers satisfying $a_1=15,27-2 a_2>0$ and $a_k=2 a_{k-1}-a_{k-2}$ for $k=3,4, \ldots, 11$.
If $\frac{a_1^2+a_2^2+\ldots+a_{11}^2}{11}=90$, then the value of $\frac{a_1+a_2+\ldots+a_{11}}{11}$ is equal to

Options:

The correct answer is:
0

Solution:
$$
\text { } \begin{aligned}
& a_k=2 a_{k-1}-a_{k-2} \\
& \Rightarrow \quad a_1, a_2, \ldots, a_{11} \text { are in AP } \\
& \therefore \frac{a_1^2+a_2^2+\ldots+a_{11}^2}{11} \\
&=\frac{11 a^2+35 \times 11 d^2+10 a d}{11}=90 \\
& \Rightarrow \quad 225+35 d^2+150 d=90 \\
& 35 d^2+150 d+135=0 \\
& \Rightarrow \quad d=-3,-\frac{9}{7}
\end{aligned}
$$
Given, $a_2 < \frac{27}{2} \therefore d=-3$ and $d \neq-\frac{9}{7}$
$$
\Rightarrow \quad \frac{a_1+a_2+\ldots+a_{11}}{11}
$$
$$
=\frac{11}{2}[30-10 \times 3]=0
$$

Topic: Mathematics, Sequences and Series, JEE Main
Let a1, a2, a3,……,a49 be in A.P. such that Σ k⁡ = 0 1 2 a4k+1=416 and a9+a43=66. If a12+a22+…+a172=140m, then m is equal to:

Options:
A 33
B 66
C 68
D 34

The correct answer is:
34

Solution:
a1+a5+ . . . +a49=416
Let common difference = d
13a1 +4d1+2+ . . . +12=416

13a1 +4d12132=416

13a1 +24d13=416
a1+24d=32 . . . (i)
a9+a43=66

a1+8d+a1+42d=66

2a1+50d=66
a1+25d=33 . . . (ii)
a1=8 ; d=1
∑ i=1 17 a i 2 =140m
⇒a2+a+d2+ .. . +a+16d2=140m
⇒17a2+d212+22+ . . . +162+2ad1+2+ . . . + 16=140m

⇒17a2+d216×17×336+2ad16×172=140m
⇒17×64+8×11×17+8×16×17=140m
⇒ m=34

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2018 (08 Apr)
Let a1, a2, a3,…an,… ,be in A.P. If a3+a7+a11+a15=72, then the sum of its first 17 terms is equal to :

Options:
A 306
B 204
C 153
D 612

The correct answer is:
306

Solution:
a3+a7+a11+a15=72
a3+a15+a7+a11=72
a3+a15=a7+a11=a1+a17
a1+a17=36
S17=172a1+a17=17×18=306

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2016 (10 Apr Online)
Let a1,a2,a3,....,an be n positive consecutive terms of an arithmetic progression. If d>0 is its common difference, then limn→∞dn1a1+a2+1a2+a3+…+1an-1+an is

Options:
A 1d
B d
C 1
D 2

The correct answer is:
1

Solution:
Given,

a1,a2,a3,....,an are terms of an A.P,

So common difference will be,

d=a2-a1=a3-a2=........=an-an-1

Now solving,

limn→∞dn1a1+a2+1a2+a3+…+1an-1+an

=limn→∞dna2-a1a2-a1+a3-a2a3-a2+…+an-an-1an-an-1

=limn→∞dn×1dan-a1

Now using the formula an=a1+n-1d we get,

=limn→∞1da1+(n-1)d-a1n

=limn→∞1dna1n+d-dn-a1nn

=1d0+d-0-0

=1d×d=1

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (06 Apr Shift 1)
Let a1,a2,a3,…. be a GP of increasing positive numbers. If the product of fourth and sixth terms is 9 and the sum of fifth and seventh terms is 24 , then a1a9+a2a4a9+a5+a7 is equal to

Options:

The correct answer is:
60

Solution:
Given,

a1,a2,a3,…. be a GP of increasing positive numbers,

And the product of fourth and sixth terms is 9

So, a4·a6=9⇒a52=9⇒a5=3

Also given the sum of fifth and seventh terms is 24,

So, a5+a7=24

⇒a5+a5r2=24

⇒1+r2=8⇒r=7

Now using a5=3⇒a1r4=3⇒a1=3r4=349

Then, a1a9+a2a4a9+a5+a7

=349×349×78+349×7×349×73×349×78+349×74+349×76

=9+27+3+21=60

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Let a1,a2,a3,…. be a G.P. of increasing positive numbers. Let the sum of its 6th and 8th terms be 2 and the product of its 3rd and 5th terms be 19. Then 6a2+a4a4+a6 is equal to

Options:
A 3
B 33
C 2
D 22

The correct answer is:
3

Solution:
Given,

a1,a2,a3,…. be a G.P. of increasing positive numbers,

And the sum of its 6th and 8th terms be 2

So, a6+a8=2

⇒ar5+ar7=2 .....1

And the product of its 3rd and 5th terms be 19,

So, a3·a5=19⇒a2·r2·r4=19

⇒ar3=13

Now from putting the value of ar3=13 in equation 1 we get,

r23+r43=2

⇒r4+r2=6

⇒r2+3r2-2=0

⇒r2=2

ar3=13⇒ar·2=13⇒ar=16

Now finding the value of 6a2+a4a4+a6, we get

=6ar+ar3ar3+ar5

=616+1313+23

=6·12·1=3

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (13 Apr Shift 2)
Let a1, a2, a3... be an A.P. with a6=2. Then, the common difference of this A.P., which maximise the product a1·a4·a5,is :

Options:
A 23
B 32
C 65
D 85

The correct answer is:
85

Solution:
Let, the first term of A.P. be a and common difference d, then we know that the nthterm of the A.P. is a+n-1d.

Then a6=a+5d=2

a=2-5d ...i

Let ∆=a1·a4·a5

⇒∆=a·(a+3d)·(a+4d)

Put the value of a from equation i,

⇒∆=2-5d·(2-5d+3d)·(2-5d+4d)

⇒∆=2-5d·(2-2d)·(2-d)

⇒∆=-2(5d3-17d2+16d-4)

For finding the maximum or minimum value of ∆

⇒d∆dd=-2(15d2-34d+16)

⇒d∆dd=-25d-8(3d-2)

Now, d∆dd=0

⇒-25d-8(3d-2)=0

⇒d=85 or d=23

And, d2∆dd2=-2(30d-34)

At d=85, d2∆dd2=-230×85-34=-28<0 and at d=23, d2∆dd2=-230×23-34=28>0

We know that if at any point the first derivative of a function is zero and the second derivative is negative, then it is the point of maxima of the function.

Hence, it is clear that ∆ is maximum when d=85.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (10 Apr Shift 2)
Let a1, a2, a3, … be an A.P. If a1+a2+…+a10a1+a2+…+ap=100p2, p≠10, then a11a10 is equal to :

Options:
A 1921
B 100121
C 2119
D 121100

The correct answer is:
2119

Solution:
Given, a1+a2+…+a10a1+a2+…+ap=S10Sp=100p2

⇒1022a1+9dp22a1+p-1d=100p2

⇒102a1+9dp2a1+p-1d=100p2

⇒2a1+9d2a1+p-1d=10p

Let, a1=a

⇒2pa+9pd=20a+10pd-10d

⇒2pa-20a=pd-10d

⇒2ap-10=dp-10

⇒2a=d

⇒ad=12

Now, a11a10=a+10da+9d

⇒a11a10=dad+10dad+9

⇒a11a10=12+1012+9

⇒a11a10=2119

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (31 Aug Shift 2)
Let a1,a2,a3,……. be an A.P. If a7=3, the product a1a4 is minimum and the sum of its first n terms is zero then n!-4ann+2 is equal to

Options:
A 3814
B 9
C 334
D 24

The correct answer is:
24

Solution:
We know the nth term of an A.P. is given by,

an=a+n-1d

Given, a7=3

⇒a+6d=3

⇒ a=3-6d

And, a1a4=aa+3d

=3-6d3-3d

=18d2-27d+9

Given product a1a4 is minimum then,

Let f(d)=18d2-27d+9

f'(d)=36d-27

Product to be minimum, f'd=0

⇒36d-27=0

⇒d=2736=34

So, a=3-92=-32

Given, Sn=0

Sn=n22a+n-1d=0

-3+n-134=0

⇒ n=5

Now n!-4an(n+2)=5!-4a35

=120-4a+34d

=120-4-32+34×34

=120+6-102=24

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Let $a_1, a_2, a_3, \ldots$ be an A.P, such that $\frac{a_1+a_2+\ldots+a_p}{a_1+a_2+a_3+\ldots+a_q}=\frac{p^3}{q^3} ; p \neq q$. Then $\frac{a_6}{a_{21}}$ is equal to:

Options:
A
$\frac{41}{11}$

B
$\frac{31}{121}$

C
$\frac{11}{41}$

D
$\frac{121}{1861}$

The correct answer is:
$\frac{31}{121}$

Solution:
$\begin{aligned} & \frac{a_1+a_2+a_3+\ldots \ldots+a_p}{a_1+a_2+a_3+\ldots \ldots+a_q}=\frac{p^3}{q^3} \\ \Rightarrow & \frac{a_1+a_2}{a_1}=\frac{8}{1} \Rightarrow a_1+\left(a_1+d\right)=8 a_1 \\ \Rightarrow \quad & d=6 a_1 \\ & \text { Now } \frac{a_6}{a_{21}}=\frac{a_1+5 d}{a_1+20 d} \\ = & \frac{a_1+5 \times 6 a_1}{a_1+20 \times 6 a_1}=\frac{1+30}{1+120}=\frac{31}{121}\end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2013 (09 Apr Online)
LetA1,A2,A3,…… be an increasing geometric progression of positive real numbers. If A1 A3 A5 A7=11296 and A2+A4=736, then, the value of A6+A8+A10 is equal to

Options:
A 43
B 33
C 37
D 48

The correct answer is:
43

Solution:
Given

A1A3A5A7=11296

Expanding using An=A1rn-1 formula we get,

⇒A1A1r2A1r4A1r6=11296

⇒A14r12=11296⇒A1r3=16 ...i

Also given ,

A2+A4=736⇒A1r+A1r3=736

⇒A1r+16=736

⇒A1r=136 ...ii

Dividing equation (i) from equation (ii) we get, r2=6

Now A6+A8+A10=A1r5+A1r7+A1r9

=A1r3r2+r4+r6 =166+36+216 =43

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2022 (28 Jun Shift 1)
Let $a_1, a_2, a_3, \ldots$ be in an arithmetic progression of positive terms.
Let $\mathrm{A}_{\mathrm{k}}=\mathrm{a}_1^2-\mathrm{a}_2^2+\mathrm{a}_3^2-\mathrm{a}_4^2+\ldots+\mathrm{a}_{2 \mathrm{k}-1}^2-\mathrm{a}_{2 \mathrm{k}}^2$.
If $\mathrm{A}_3=-153, \mathrm{~A}_5=-435$ and $\mathrm{a}_1^2+\mathrm{a}_2^2+\mathrm{a}_3^2=66$, then $\mathrm{a}_{17}-\mathrm{A}_7$ is equal to______

Options:

The correct answer is:
910

Solution:
$\begin{aligned} & \mathrm{d} \rightarrow \text { common diff. } \\ & \mathrm{A}_{\mathrm{k}}=-\mathrm{kd}[2 \mathrm{a}+(2 \mathrm{k}-1) \mathrm{d}] \\ & \mathrm{A}_3=-153 \\ & \Rightarrow 153=13 \mathrm{~d}[2 \mathrm{a}+5 \mathrm{~d}] \\ & 51=\mathrm{d}[2 \mathrm{a}+5 \mathrm{~d}]...(1) \\ & \mathrm{A}_5=-435 \\ & 435=5 \mathrm{~d}[2 \mathrm{a}+9 \mathrm{~d}] \\ & 87=\mathrm{d}[2 \mathrm{a}+9 \mathrm{~d}] \\ & (2)-(1) \\ & 36=4 \mathrm{~d}^2 \\ & \mathrm{~d}=3, \mathrm{a}=1 \\ & \mathrm{a}_{17}-\mathrm{A}_7=49-[-7.3[2+39]]=910\end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (05 Apr Shift 1)
Let $A_{1}, A_{2}, A_{3}$ be regions in the $x y$-plane defined by
$$
\begin{array}{l}
A_{1}=\left\{(x, y): x^{2}+2 y^{2} \leq 1\right\} \\
A_{2}=\left\{(x, y):|x|^{3}+2 \sqrt{2}|y|^{3} \leq 1\right\}
\end{array}
$$
$A_{3}=\{(x, y): \max (|x|, \sqrt{2}|y|) \leq 1$. Then

Options:
A $\mathrm{A}_{1} \supset \mathrm{A}_{2} \supset \mathrm{A}_{3}$
B $A_{3} \supset A_{1} \supset A_{2}$
C $\mathrm{A}_{2} \supset \mathrm{A}_{3} \supset \mathrm{A}_{2}$
D $A_{3} \supset A_{2} \supset A_{1}$

The correct answer is:
$A_{3} \supset A_{2} \supset A_{1}$

Solution:

Topic: Mathematics, Area Under Curves, JEE Main
Let A1,A2,A3,….. be squares such that for each n⩾1, the length of the side of An equals the length of diagonal of An+1. If the length of A1 is 12 cm, then the smallest value of n for which area of An is less than one, is

Options:

The correct answer is:
9

Solution:
Let an be the side length of An.

So, an=2an+1,a1=12

⇒ an=12×12n-1

Now, an2<1⇒1442n-1<1

⇒ 2n-1>144

⇒ n-1≥8

⇒ n≥9

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (25 Feb Shift 1)
Let $a_1, a_2, a_3, \ldots$ be terms of an A.P. If $\frac{a_1+a_2+\cdots a_p}{a_1+a_2+\cdots+a_q}=\frac{p^2}{q^2}, p \neq q$, then $\frac{a_6}{a_{21}}$ equals

Options:
A
$\frac{41}{11}$

B
$\frac{7}{2}$

C
$\frac{2}{7}$

D
$\frac{11}{41}$

The correct answer is:
$\frac{11}{41}$

Solution:
$$
\begin{aligned}
& \frac{\frac{p}{2}\left[2 a_1+(p-1) d\right]}{\frac{q}{2}\left[2 a_1+(q-1) d\right]}=\frac{p^2}{q^2} \Rightarrow \frac{2 a_1+(p-1) d}{2 a_1+(q-1) d}=\frac{p}{q} \\
& \frac{a_1+\left(\frac{p-1}{2}\right) d}{a_1+\left(\frac{q-1}{2}\right) d}=\frac{p}{q}
\end{aligned}
$$
For $\frac{a_6}{a_{21}}, p=11, q=41 \rightarrow \frac{a_6}{a_{21}}=\frac{11}{41}$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2006
Let a1,a2,a3 be three positive numbers which are in geometric progression with common ratio r. The inequality a3>a2+2a1 holds true if r is equal to

Options:
A 2
B 1.5
C 0.5
D 2.5

The correct answer is:
2.5

Solution:
a2=a1r,a3=a1r2
a3>a2+2a1
⇒a1r2>a1r+2a1
⇒r2-r-2>0
⇒r-2r+1>0
⇒r<-1 or r>2

Topic: Mathematics, Sequences and Series, JEE Main
Let a1,a2,...,a10 be 10 observations such that ∑k=110ak=50 and ∑∀k<jak·aj=1100. Then the standard deviation of a1,a2,…,a10 is equal to :

Options:
A 5
B 5
C 10
D 115

The correct answer is:
5

Solution:
Given: ∑k=110ak=50

⇒a1+a2+…+a10=50 ...i

⇒a1+a2+…+a102=2500

⇒∑i=110ai2+2∑k<jakaj=2500

⇒∑i=110ai2+2×1100=2500

⇒∑i=110ai2=2500-21100

⇒∑i=110ai2=300

Standard deviation, σ=∑ai2n-∑ain2

⇒σ=∑ai210-∑ai102

⇒σ=30010-50102

⇒σ=30-25=5

Topic: Mathematics, Statistics, JEE Main, JEE Main 2024 (27 Jan Shift 1)
Let $a_{1}, a_{2}, \ldots, a_{10}$ be a G.P. If $\frac{a_{3}}{a_{1}}=25,$ then $\frac{a_{9}}{a_{5}}$ equals :

Options:
A $5^{4}$
B $4\left(5^{2}\right)$
C $5^{3}$
D $2\left(5^{2}\right)$

The correct answer is:
$5^{4}$

Solution:
Let $a_{1}=a, a_{2}=a r, a_{3}=a r^{2} \ldots a_{10}=a r^{9}$

where $r=$ common ratio of given G.P.

Given, $\frac{a_{3}}{a_{1}}=25$

$\Rightarrow \frac{a r^{2}}{a}=25$

$\Rightarrow r=\pm 5$

Now, $\frac{a_{9}}{a_{5}}=\frac{a r^{8}}{a r^{4}}=r^{4}=(\pm 5)^{4}=5^{4}$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (11 Jan Shift 1)
Let a1,a2,…,a10 be an A.P. with common difference -3 and b1, b2,…,b10 be a G.P. with common ratio 2. Let ck=ak+bk, k=1,2,…,10. If c2=12 and c3=13, then ∑k=110ck is equal to ______.

Options:

The correct answer is:
2021

Solution:
Ck=ak+bk

where ak=a1+(k-1)(-3) and bk=b12k-1

Ck=a1+(k-1)(-3)+b12k-1

∑k=110ck=∑k=110ak+∑k=110bk

C2=a1-3+b1(2)=12 (1)....

C3=a1-6+4 b1=13 (2)......

by solving (1) & (2)
2b1-3=1⇒b1=2 and a1=11

∑k=110ck=10222+9-3+2210-12-1

=-25+2046=2021

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (26 Aug Shift 2)
Let a1, a2, …, a21 be an A.P. such that ∑n=1201anan+1=49. If the sum of this A.P. is 189, then a6a16 is equal to :

Options:
A 57
B 48
C 36
D 72

The correct answer is:
72

Solution:
∑n=1201anan+1=49 [let common difference of AP is d=an+1-an

=1d∑n=1201an-1an+1⇒1d1a1-1a2+1a2-1a3+….1a20-1a21=49

⇒1d1a1-1a21=49

⇒1da21-a1a1·a21=49⇒20dda1a21=49⇒a1a21=45

a1a1+20d=45 ...(1)

Now sum of first 21 terms=2122a1+20d=189⇒a1+10d=9 ...(2)

by using equation (1) and (2) we get a1=3, d=35 otherwise a1=15, d=-35

So, a6a16=a1+5da1++15d=72

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Let a1, a2, …, a30 be an A.P., S= ∑i=130ai and T= ∑i=115a(2i-1). If a5=27 and S-2T=75, then a10 is equal to:

Options:
A 52
B 47
C 42
D 57

The correct answer is:
52

Solution:
Let common difference =d
∴S= ∑i=130ai=3022a1+29 d
∴S=15 2a1+29 d ........ (i)
T=1522a1+14 . 2d
T=15 a1+14d ...........(ii)
a1+4d=27 ......... (iii)
S-2T=75
⇒152a1+29 d-30 a1+14d =75
⇒15 d29-28= 75
⇒d=5
∴ from (iii) a1=7
∴ a10=a1+9d=7+45=52

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Let a,1a2,…,an be a given A.P. whose common difference is an integer and Sn=a1+a2+…+an. If a1=1,an=300 and 15≤n≤50, then the ordered pair Sn-4,an-4 is equal to:

Options:
A (2490,249)
B (2480,249)
C (2480,248)
D (2490,248)

The correct answer is:
(2490,248)

Solution:
an=a1+(n-1)d

300=1+(n-1)d
⇒d=299(n-1)=13×23(n-1)= integer
so n-1=±13,±23,±299,±1
⇒ n=14,-12,24,-22,300,-298,2,0
But n∈[15, 50] ⇒n=24 ⇒d=13
Hence,

Sn-4=S20=2022(1)+(20-1)(13)

⇒Sn-4=2490
And,

an-4=a20=a1+19d

=1+19×13

=248

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (04 Sep Shift 2)
Let a1,a2,……,an be in A.P. If a5=2a7 and a11=18, then 121a10+a11+1a11+a12+…..1a17+a18 is equal to _____ .

Options:

The correct answer is:
8

Solution:
Let the common difference of an A.P. be d, then

2a7=a5 (given)

⇒2a1+6d=a1+4d

⇒a1+8d=0 …1

And,

a11=18

⇒a1+10d=18 …2

Solving 1 and 2, we get

a1=-72,d=9

So,

a18=a1+17d=-72+153=81

a10=a1+9d=9

Now,

121a10+a11+1a11+a12+…..1a17+a18

=12a11-a10 d+a12-a11 d+……a18-a17 d

=12a18-a10d=129-39=8

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Let $a_{1}, a_{2}, \ldots a_{n}$ be $n$ nonzero real numbers, of which $p$ are positive and remaining are negative. The number of ordered pairs $(\mathrm{j}, \mathrm{k}), \mathrm{j} < \mathrm{k}$, for which $\mathrm{a}_{\mathrm{j}} \mathrm{a}_{\mathrm{k}}$ is positive, is $55 .$ Similarly, the number of ordered pairs $(\mathrm{j}, \mathrm{k}), \mathrm{j} < \mathrm{k}$, for which a $a_{k}$ is negative is 50 . Then the value of $p^{2}+(n-p)^{2}$ is

Options:
A 629
B 325
C 125
D 221

The correct answer is:
125

Solution:
${ }^{\mathrm{P} C}_{2}+{ }^{n-p} \mathrm{C}_{2}=55$
$\frac{\mathrm{p}(\mathrm{p}-1)}{2}+\frac{(\mathrm{n}-\mathrm{p})(\mathrm{n}-\mathrm{p}-1)}{2}=55$ .........(i)
Also,$p(n-p)=50$.........(ii)
Put in (i)
$\begin{array}{l}
p(p-1)+\frac{50}{p}\left(\frac{50}{p}-1\right)=110 \\
p^{2}-p+\left(\frac{50}{p}\right)^{2}-\frac{50}{p}=110 \\
p^{2}-p+\left(\frac{50}{p}\right)^{2}-\frac{50}{p}=110 \\
\left(p+\frac{50}{p}\right)^{2}-100-\left(p+\frac{50}{p}\right)=110 \\
t^{2}-t-210=0 \\
t=15 \text { or }-14 \text { (not true) } \\
\therefore \quad p+\frac{50}{p}=15
\end{array}$
$\therefore$ To find $p^{2}+(n-p)^{2}=p^{2}+\left(\frac{50}{p}\right)^{2}$
$\begin{array}{l}
=\left(p+\frac{50}{p}\right)^{2}-100 \\
=125 \text { (using (iii)) }
\end{array}$

Topic: Mathematics, Permutation Combination, JEE Main
Let $a_{1}, a_{2}, \ldots a_{n}$ be $n$ nonzero real numbers, of which $p$ are positive and remaining are negative. The number of ordered pairs $(\mathrm{j}, \mathrm{k}), \mathrm{j} < \mathrm{k}$, for which $\mathrm{a}_{\mathrm{j}} \mathrm{a}_{\mathrm{k}}$ is positive, is $55 .$ Similarly, the number of ordered pairs $(\mathrm{j}, \mathrm{k}), \mathrm{j} < \mathrm{k}$, for which a $a_{k}$ is negative is 50 . Then the value of $p^{2}+(n-p)^{2}$ is

Options:
A 629
B 325
C 125
D 221

The correct answer is:
125

Solution:
${ }^{\mathrm{P} C}_{2}+{ }^{n-p} \mathrm{C}_{2}=55$
$\frac{\mathrm{p}(\mathrm{p}-1)}{2}+\frac{(\mathrm{n}-\mathrm{p})(\mathrm{n}-\mathrm{p}-1)}{2}=55$ .........(i)
Also,$p(n-p)=50$.........(ii)
Put in (i)
$\begin{array}{l}
p(p-1)+\frac{50}{p}\left(\frac{50}{p}-1\right)=110 \\
p^{2}-p+\left(\frac{50}{p}\right)^{2}-\frac{50}{p}=110 \\
p^{2}-p+\left(\frac{50}{p}\right)^{2}-\frac{50}{p}=110 \\
\left(p+\frac{50}{p}\right)^{2}-100-\left(p+\frac{50}{p}\right)=110 \\
t^{2}-t-210=0 \\
t=15 \text { or }-14 \text { (not true) } \\
\therefore \quad p+\frac{50}{p}=15
\end{array}$
$\therefore$ To find $p^{2}+(n-p)^{2}=p^{2}+\left(\frac{50}{p}\right)^{2}$
$\begin{array}{l}
=\left(p+\frac{50}{p}\right)^{2}-100 \\
=125 \text { (using (iii)) }
\end{array}$

Topic: Mathematics, Permutation Combination, KVPY, KVPY 2017 (5 Nov SA)
Let A1, A2 and A3 be the regions on R2 defined by

A1=x,y:x≥0,y≥0,2x+2y-x2-y2>1>x+y

A2=x,y:x≥0,y≥0,x+y>1>x2+y2

A3=x,y:x≥0,y≥0,x+y>1>x3+y3

Denote by A1, A2 and A3 the areas of the regions A1, A2 and A3 respectively. Then

Options:
A A1>A2>A3
B A1>A3>A2
C A1=A2<A3
D A1=A3>A2

The correct answer is:
A1=A2<A3

Solution:
Area of region defined by

A1=(x,y):x≥0,y≥0,2x+2y-x2-y2>1>x+y is A1.

Similarly, A2 is the area of region defined by

A2=x,y:x≥0,y≥0,x+y>1>x2+y2

Here, A1=A2

and area A3 is the area of region defined by A3=x,y:x≥0,y≥0,x+y>1>x3+y3
then A3>A2=A1

Topic: Mathematics, Area Under Curves, JEE Main
Let A=1a3001b3c301 where a,b,câR. If the sum of all non-real roots of the equation A-xI=0 is k-mabc, âk,mâZ, then the value of k+m is equal to

Options:

The correct answer is:
3

Solution:
A-nI=1-na3001-nbc301-n=0
⇒1-n1-n2-a3-b3c3=0
⇒-n-13+a3b3c3=0
⇒n-13=abc3
⇒n=1+abc,1+abcw,1+abcw2
Sum of non real roots =2+abcw+w2
=2-abc
k=2,m=1
⇒k+m=3

Topic: Mathematics, Determinants, JEE Main
Let A1 and A2 be two arithmetic means and G1, G2 and G3 be three geometric means of two distinct positive numbers. Then G14+G24+G34+G12G32 is equal to

Options:
A A1+A22G1G3
B 2A1+A2G1G3
C A1+A2G12G32
D 2A1+A2G12G32

The correct answer is:
A1+A22G1G3

Solution:
Let the two numbers are a, b

⇒a,A1,A2,b are in AP.

⇒b=a+4-1d

⇒d=b-a3

⇒A1=a+b-a3=2a+b3

⇒A2=a+b-a3·2=a+2b3

Similarly a,G1,G2,G3,b are in GP.

⇒b=ar5-1

⇒r=ba14

⇒G1=aba14

⇒G2=aba24

⇒G3=aba34

=G14+G24+G34+G12·G32

⇒a4·ba+a4·b2a2+a4·b3a3+a4·b2a2

=ba3+b2a2+b3a+a2b2

=aba2+b2+2ab=aba+b2

⇒A1+A22·G1G3=a+b2·ab

Hence this is the correct option.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (15 Apr Shift 1)
Let a→=1,b→=4 and a→×r→+b→=r→. If the projection of r→ along a→ is 2, then the projection of r→ along b→ is

Options:

The correct answer is:
2.5

Solution:
aâ=1,bâ=4,rââa^=2
aâÃrâ+b^=râ
Taking dot with aâ
0+bââaâ=rââaâ
bââaâaâ=rââaâaâ=2
âaââbâ=2 ...(i)
âaâbâcosâ¡Î¸=2
âcosâ¡Î¸=12âÎ¸=60o (angle between aâ and bâ )
Taking dot with bâ
aâÃrââbâ+bââbâ=rââbâ
-râaâbâ+16=rââbâ ...(ii)
Taking dot with aâÃbâ
aâÃrââaâÃbâ+0=râaâbâ
aââaâaââbârââaârââbâ=16-rââbâ (using (ii))
1rââbâ-2Ã2=16-rââbâ (using (i))
ârââbâ=10
Hence, the projection of râ on bâ is 52

Topic: Mathematics, Vector Algebra, JEE Main
Let a>1 be a constant. If f:A→A and x, y∈f satisfy ax+ay=a, then A=

Options:
A (0, a]
B [0, a]
C (-∞, 1)
D (-∞, a+1)

The correct answer is:
(-∞, 1)

Solution:
Given ax+ay=a

⇒ay=a-ax

Taking log with base a to both the sides, we get

⇒logaay=logaa-ax

Using logmn=nlogm,

⇒ylogaa=logaa-ax

Using logaa=1,

⇒y=logaa-ax

The domain of the function A is defined if a-ax>0 & a≠1

⇒ax<a

Again, taking log with base a, both the sides, we get

logaax<logaa

⇒xlogaa<1

⇒x<1.

Hence, the domain A of the function is -∞, 1.

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2021 (05 Aug Shift 2)
Let A1 be the area of the region bounded by the curves y=sinx,y=cosx and y-axis in the first quadrant. Also, let A2 be the area of the region bounded by the curves y=sinx,y=cosx, x-axis and x=π2 in the first quadrant. Then,

Options:
A 2A1=A2 and A1+A2=1+2
B A1:A2=1:2 and A1+A2=1
C A1:A2=1:2 and A1+A2=1
D A1=A2 and A1+A2=2

The correct answer is:
A1:A2=1:2 and A1+A2=1

Solution:
A1=∫0π/4cosx-sinxdx

A1=sinx+cosx0π/4=2-1

A2=∫0π/4sinxdx+∫π/4π/2cosxdx

=-cosx0π/4+sinxπ/4π/2

A2=1-12+1-12=2-2

A1:A2=1:2, A1+A2=1

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2021 (26 Feb Shift 2)
Let A=1+i1-i0 where i=-1. Then, the number of elements in the set n∈{1,2,….,100}:An=A is

Options:

The correct answer is:
25

Solution:
We have A=1+i1-i0

A2=A·A=1+i1-i01+i1-i0=i1+i-i+1-i

A3=A2·A=i1+i1-i-i1+i1-i0=0i11-i

A4=A3·A=0i11-i1+i1-i01001=I

∴ A4=I

So, A5=A4·A=I·A=A

A6=A4·A2=I·A2=A2 and so on

∵A1=A5=A9=.....=A97=A

Hence, possible values of n, such that An=A

=1,5,9,…,97

Clearly, above sequence is in A.P. where

a=1, d=4 & tn=97 ⇒a+n-1d=97

⇒ 1+n-14=97 ⇒ n=25

∴ The number of elements in the given set =25.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2022 (28 Jun Shift 2)