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96. Given, series is $2*4+4*6+6*8+\dots +n$ terms.

So, a₂₀ = ( 20th term of A.P. 2, 4, 6 ….) ( 20th term of A.P. 4,6,8 ………)

i.e. a = 2, d = 4 -2 = 2 i.e. a =4, d = 6- 4 = 2

= [2 + (20- 1)2] [4 + (20-1)2]

= [2+19*2] [4 +19 *2]

= (2+38) (4+38)

= 40*42 = 1680

Some of the top recruiters for D. Y. Patil College of Engineering and Technology placement are listed below:

The documents that are important for admissions to MMS courses at TIMSR Mumbai at the time of applying for the course and at the time of admission confirmation  are as follows:

• Class 10 & 12 Marksheet 
• Graduation Marks all semesters
• Entrance Test Scorecards 
• College Leaving/ Transfer Certificate 
• Migration Certificate 
• Domicile Certificate
• Character Certificate, etc. 

95. (i) Sn = 5+55+555+…………… upto n term

=5(1+11+111+ …………….upto n term )

Multiplying and dividing by 9 we get

$=\frac{5}{9}(9+99+999+\dots upto\mathrm{nterms})$

 

$=\frac{5}{9}\left[\right(10+10^2+10^3+?$ upto\mathrm{nterms}$$
$-n\times 1]$

$=\frac{5}{9}\left[\frac{10\left(10^n-1\right)}{10-1}-n\right]$ { $?10+10^2+10^3+\dots \mathrm{ntermsisaG.Pof}a=10,r=10>1\mathrm{nterms}$ }

$=\frac{5}{9}\times \left[\frac{10}{9}\left(10^n-1\right)-n\right]$

$=\frac{50}{81}\left(10^n-1\right)-\frac{5n}{9}$

(ii) $S_n=0.6+0.66+0.666+\dots n$

$=6[0.1+0.11+0.111+?uptonterms$$]$

$=\frac{6}{9}[0.9+0\cdot 99+0.999+?\mathrm{<math><mrow><mo>uptonterms</mo></mrow></math>}]$ {multiplying and dividing by 9}

$=\frac{6}{9}[(1+1+1\dots n)-(0.1+0.01+0.001+\dots \mathrm{<math><mrow><mo>uptonterms</mo></mrow></math>})]$

$=\frac{6}{9}\left[n-\left(\frac{1}{10}+\frac{1}{10}\times \frac{1}{10}+\frac{1}{10}\times {\left(\frac{1}{10}\right)}^2+?\mathrm{<math><mrow><mo>uptonterms</mo></mrow></math>}\right)\right]$

$=\frac{6}{9}\left[n-\frac{\frac{1}{10}\left(1-{\left(\frac{1}{10}\right)}^n\right)}{1-\frac{1}{10}}\right]$

$=\frac{6}{9}\left[n-\frac{1-{\left(\frac{1}{10}\right)}^n}{10-1}\right]$

$=\frac{6}{9}\left[x-\frac{1}{9}\left(1-\frac{1}{10^n}\right)\right]$

$=\frac{6}{9}\times \frac{1}{9}\left[9x-\left(1-\frac{1}{10^n}\right)\right]$

$=\frac{2}{27}\left[9n-\left(1-\frac{1}{10^n}\right)\right]$

$=\frac{2}{27}\left[9n-1+\frac{1}{10^n}\right]$

Yes, AURO University indeed offers CUET scholarships for undergraduate programmes. Tthe university provides a certain percentage of scholarship depedning on the score secured by the candidates in CUET (best of five subjects). Check out the table below to learn about AURO University CUET scholarships:

Note: This information is taken from official sources and is subject to change

94. As a, b, c are in A.P. we can write,

$b-a=c-b$

$\Rightarrow b+b=a+c$

$\Rightarrow 2b=a+c$ I

As b, c, d are in G.P. we can write,

$\frac{c}{b}=\frac{d}{c}$

$\Rightarrow c^2=bd$ II

And ar $\frac{1}{c},\frac{1}{d},\frac{1}{e}$ are in A.P. we can write,

$\Rightarrow \frac{1}{d}-\frac{1}{c}=\frac{1}{e}-\frac{1}{d}$

$\Rightarrow \frac{1}{d}+\frac{1}{d}=\frac{1}{c}+\frac{1}{e}$

$\Rightarrow \frac{2}{d}=\frac{e+c}{ce}$

$\Rightarrow \frac{d}{2}=\frac{ce}{e+c}$

$\Rightarrow d=\frac{2ce}{c+e}$ …………. III

Now, $c^2=Bd$ from II

$\Rightarrow =\frac{a+c}{2}\times \frac{2ce}{c+c}$ {from 1 and 3}

$\Rightarrow c^2=\frac{(a+c)ce}{c+e}$

$\Rightarrow c=\frac{(a+c)e}{(c+e)}$

$\Rightarrow c(c+e)=(a+c)e$

$\Rightarrow c^2+ce=ac+ce$

$\Rightarrow c^2=ae$

$\Rightarrow \frac{c}{a}=\frac{e}{c}$

i.e., a, c and e are in G.P.

A total of 271 job offers have been made in the D. Y. Patil College of Engineering and Technology placement in 2024. Refer to the table for more clarity on job offers provided during the placements in 2022-2024:

The D. Y. Patil College of Engineering and Technology has had a strong placement record since its inception. Check out the tabulated data given below to know more about D. Y. Patil College of Engineering and Technology placement in 2022-2024:

Indian Institute of technology Ropar is one of the premier Engineering institutes in India, with high ranking performance in various indices. The courses at IIT Ropar, like BTech, MTech, MSc, etc., are offered by various faculty departments at the campus. The different IIT Ropar departments are listed below:

1. Artificial Intelligence and Data Engineering
2. Chemical
3. Chemistry
4. Civil
5. CSE
6. Electrical
7. HSS
8. Mathematics
9. Mechanical
10. Metallurgical and Materials Engineering
11. Physics

A total of 222 students secured jobs in the D. Y. Patil College of Engineering and Technology placement in 2024. Refer to the table for more clarity:

Although IIT Ropar is one of the newer IIT institutions established by the Government of India, the fact remains that admissions to the institute still remain particularly competitive among Indian students. The BTech course is often sought out by various students to receive quality technical education and to prepare them with the right set of skills. Of particular demand is the CSE branch of BTech in the various IITs across the nation. Students must prepare for their studies accordingly.

D. Y. Patil College of Engineering and Technology provides good  placement opportunities to its graduating students. The students are given excellent placement support every year placing them in top companies. Check out the table below to know the key highlights of the D. Y. Patil College of Engineering and Technology placement in 2022-2024:

IIT Ropar has a large number of prestigious engineering courses in different specializations that attract students from all around. The flagship BTech program is one of the sought-after courses at the campus. The BTech programme is offered by the IIT Ropar faculty with branches like Computer Science, Electrical Engineering, Mechanical Engineering, Chemical Engineering, and several others. Various top-name companies visit the campus during placements season to recruit the best of talent from BTech students.

IIT Ropar is a renowned college and offers engineering courses to candidates. These are- BTech, MTech, PhD, etc. The placements are also good and the campus also has great infrastructure. 

92. Given, ab are roots of $x^2-3x+p=0$

and c & d are roots of $x^2-12x+q=0$

So, $\Rightarrow a+b=\frac{-\left(-3\right)}{1}$ and $ab=\frac{P}{1}$

a + b = +3 ………….I and ab = P…………….II { $?$ sum of roots = $\frac{-B}{A}$ , Product of roots = $\frac{-C}{A}$ }

Similarly, c + d $=\frac{-(-12)}{1}$ and $cd=\frac{q}{1}$

c + d = 12 ……….III ad cd = q (4)

As a, b, c, d from a G.P and if r be the common ratio

a = a

b = ar

c = ar²

d = ar³

So, from equation, (1),

$a+b=3\Rightarrow a+ar=3\Rightarrow a(1+r)=3$ (5)

And $c+d=12\Rightarrow a{r}^2+a{r}^3=12\Rightarrow a{r}^2(1+r)=12$ (6)

Dividing equation (6) and (5) we get,

$\frac{a{r}^2(1+r)}{a(1+r)}=\frac{12}{3}$

 r² = 4

Now, L.H.S. $=\frac{q+p}{q-p}=\frac{cd+ab}{cd-ab}$ {from (4) and (5)}

$=\frac{a{r}^2\times a{r}^3+a\times ar}{a{r}^2+a{r}^3-a\times ar}$

$=\frac{a^{2}r\left(r^4+1\right)}{a^{2}r\left(r^4-1\right)}$

$=\frac{{\left(r^2\right)}^2+1}{{\left(r^2\right)}^2-1}=\frac{4^2+1}{4^2-1}=\frac{16+1}{16-1}=\frac{17}{15}$ = R.H.S.

D. Y. Patil College of Engineering and Technology has been providing successful on campus placements over the years. Check out the table below to know the key highlights of the D. Y. Patil College of Engineering and Technology placement in 2022-2024:

Yes, AURO University does offer scholarships. The university provides several scholarship schemes for its students who meet the eligibility criteria for the desired scheme. Candidates can avail of the schemes by applying for the same and completing the entire process. Some of the scholarships offered at the university are as follows:

• AURO Merit Scholarship
• Sports Scholarship
• The university's student performance-based scholarship programmes
• Need-based financial aid
• AURO Meritorious Scholarship

91. Let r be the common ratio of the G.P.

Then, a, b, c, d a, ar, ar², ar³

So, $a^n+b^n=a^n+{(ar)}^n=a^n+a^n{r}^n=a^n\left(1+r^n\right)-(1)$

$b^n+c^n={(ar)}^n+{\left(a{r}^2\right)}^n=a^n{r}^n+a^n{r}^{2n}=a^n{r}^n\left(1+r^n\right)$ (2)

$c^n+d^n={\left(a{r}^2\right)}^n+{\left(a{r}^3\right)}^n=a^n{r}^{2n}+a^n{r}^{3n}=a^n{r}^{2n}\left(1+r^n\right)$ (3)

Hence, $\frac{b^n+c^n}{a^n+b^n}=\frac{a^n{r}^n\left(1+r^n\right)}{a^n\left(1+r^n\right)}=r^n$ {from (2) and (1)}

and $$ $\frac{c^n+d^n}{b^n+c^n}=\frac{a^n{r}^{2n}\left(1+r^2\right)}{a^n{r}^n\left(1+r^2\right)}=r^n$ {from (3) and (2)}

i.e., $\frac{b^n+c^n}{a^x+b^n}=\frac{c^n+d^n}{b^x+c^n}$

$\therefore a^n+b^n,b^n+c^n,c^n+d^n$ are in A.P