Ask & Answer: India's Largest Education Community

Khalifa University has a competitive acceptance rate of around 9%. This is means only 9 out of 100 students gets admission in Khalifa university. 

Students from all over the world apply to UCR and the acceptance rate here stands at 76% which is a jump from the last year. This acceptance rate can be considered good but to take admission at this university, Indian students are required to meet certainf UCR admission requirements that are mentioned below: 

UG Requirements:

• Subject-specific requirements
• English language requirements
• GPA scores
• ACT/SAT (optional)
• Transcripts

Graduate Requirements:

• Academic transcripts
• Resume/CV
• Statement of purpose
• Work experience
• Essay
• GPA scores
• Letter of Recommendation
• GRE/GMAT (optional)
• English language requirements

31. Since the man starts paying installmentas? 100 and increases? 5 every month.

a=100

d=5.

So, the A.P is 100,105,110,115, ….

? Amount of 30thinstallment=30th term of A.P =a₃₀

=a+ (30 – 1)d

=100+29 × 5.

=100+145

= ?245

i.e., He will pay? 245 in his 30th instalment.

TSEC Mumbai has a good number of seats available for undergraduate engineering courses. As per official sources, the institute offers around 540 seats for admission to its BE courses. These seats are distributed among the specialisations offered by the institute. However, the distribution of seats among various specialisations may vary, depending on the institute's specific allocation.

The University of California Riverside Campus does accept the international students and these students can apply for the same by using the official admission portal for PG courses and UCAS for UG courses. It is crucial to meet the admission requirements in order to be a part of University of California Riverside Campus admissions.  

29. Given,

A.M between a and b= $\frac{a+b}{2}$

And $\frac{a^n+b^n}{a^{n-1}+b^{n-1}}=\frac{a+b}{2}$

2 [aⁿ+bⁿ]= (a+b) (a^{n – 1}+b^{n – 1})

2aⁿ+2bⁿ=a^{n – 1}.a+ab^{n – 1}+ba^{n – 1} + b^{n – 1}.b

2aⁿ+2bⁿ=a^{n – 1 + 1}+ ab^{n – 1}+ba^{n – 1} + b^{n – 1 + 1}

2aⁿ+2bⁿ=aⁿ+ ab^{n – 1}+b^{n – 1} + bⁿ

2aⁿ – aⁿ – ban ^{– 1} = ab^{n – 1} + bⁿ– 2bⁿ

aⁿ – ba^{n – 1}=ab^{n – 1} – bn

a^{n – 1} [a – b]=b^{n – 1} [a – b] [?   aⁿ=a^{n – 1+1}=a^{n – 1} .a¹]

a^{n – 1}=b^{n – 1}.

$\frac{a^{n-1}}{b^{n-1}}=1$ .

${\left(\frac{a}{b}\right)}^{n-1}={\left(\frac{a}{b}\right)}^0$ [? a⁰=1].

n – 1=0

n=1.

PES University offers BArch for the duration of 5 years with an annual intake of 40 students. The University offers BArch based on the minimum aggregate of 45% in Class 12 and accepted entrance exams such as KCET, NATA, NTA or other equivalent exams. PESU Bangalore offers BArch with a total tuition fees worth INR 20 Lakh. 

The list of documents that candidates must keep a scanned copy of with themselves inorder to fill the application form successfully are: 

Class 10 marksheet

Class 12 marksheet

Graduation marksheet

Transfer Certificate fromt the last passing institute

Candidates with these documents can successfully submit the application form and register for admission. 

28. Let A₁, A₂, A₃, A₄and A₅ be five numbers between 8 and 26.

So, the A.P is 8, A₁, A₂, A₃, A₄, A₅, 26. with n=7 terms.

Also, a = 8.

l=26, last term.

a+ (7 – 1)d=26

8+6d=26

6d=26 – 8

d= $\frac{18}{6}$

d=3

So,

A₁=a+d=8+3=11

A₂=a+2d=8+2 × 3=8+6=14

A₃=a+3d=8+3 × 3=8+9=17.

A₄=a+9d=8+4 × 3=8+12=20

A₅=a+5d=8+5 × 3=8+15=23.

Hence the reqd. five nos are 11,14,17,20 and 23.

TSEC Mumbai has received rankings by many ranking bodies, including Times, etc. During 2024, it was ranked #42in the Times 2024 'BE/BTech' category. Observing the past three years Times BE ranking trend, the institute was ranked continuously at #42.

In order to make a choice between BE/BTech courses offered by Thadomal Shahani Engineering College and Dwarkadas J. Sanghvi College of Engineering, students can compare the options based on significant factors such as fees and Shiksha rating. To help, presented below is a comparison between the BE/BTech courses at Thadomal Shahani Engineering College and Dwarkadas J. Sanghvi College of Engineering based on fees and ratings:

Note: The abovementioned information is as per official sources. However, it is subject to change. Hence, it is indicative.

27. Given,

Sum of n terms of AP, S_n=3n²+5n

Put n=1,

S₁=3 × 1²+5 × 1=3+5=8=a₁ (?   S₁=sum of 1* terms of AP)

Put n=2,

S₂=3 × 2²+5 × 2=3 × 4+10=12+10 =22

=a₁+a₂ (?   Sum of first two term of A.P)

So,  a₁+a₂=22

8 +a₂=22

a₂=22 – 8

a₂=14

? First term,  a=a₁=8

Common difference,  d=a₂ – a₁=14 – 8=6

Now, given that, a_m = 164.

a+ (m – 1)d=164.

8+ (m – 1)6=164.

(m – 1)6=164 – 8.

m – 1= $\frac{156}{6}$

m=26+1

m=27.

26. Let a and d be the first term and common difference of A.P.

Then, $\frac{\mathrm{Sum\; of\; M\; terms\; of\; A.P.}}{\mathrm{SumofntermsofA.P.}}=\frac{m^2}{n^2}$

$\frac{\frac{m}{2}[2a+(m-1)d]}{\frac{m}{2}[2a+(n-1)d]}=\frac{m^2}{n^2}$

$\frac{m[2a+(m-1)d]}{n[2a+(n-1)d]}=\frac{m^2}{n^2}$

Dividing both sides by m/n we get,

$\frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m}{n}$

[2a+(n – 1)d]n=m[2a+(n – 1)d].

2an+dmn – dn=2am+dmn – dm.

2an – 2am=dn – dm+2mn – dmn

2a(n – m)=d(n – m)

d= $\frac{2a(n-m)}{(n-m)}$

d=2a.

Now, Ratio of m^th  and n^thterm

= $\frac{a+(m-1)d}{a+(n-1)d}$

Putting d=2a in the above we get,

ratio of mth and nth term = $\frac{a+(m-1)2a}{a+(n-1)2a}$

= $\frac{a+2am-2a}{a+2an-2a}$

= $\frac{2am-a}{2an-a}$

= $\frac{a[2m-1]}{a[2n-1]}$

= $\frac{2m-1}{2n-1}$ .

The four types of IELTS exams are: IELTS Academic, IELTS General Training, IELTS for UKVI and IELTS for UKVI Life Skills. One should confirm the university or institutions about the types of IELTS exam they accepts. On the basis of IELTS types requirements, candidates should register for IELTS exam. 

25. Let e and d the first term and common difference of an AP.

Then,

Sum of first p terms, S_p=a.

$\frac{p}{2}[2e+(p-1)d]=a$

$\frac{1}{2}[2e+(p-1)d]=\frac{a}{p}$ .

Similarly, S_q=b

$\frac{q}{2}[2e+(q-1)d]=b$

$\frac{1}{2}[2e+(q-1)d]=\frac{b}{q}$ ----------------(2)

And. S_r=C

$\frac{r}{2}[2e+q(r-1)d]=c$

$\frac{1}{2}[2c+(r-1)d]=\frac{c}{r}$ -----------------(3)

So, L.H.S. $\frac{a}{p}(q-r)+\frac{b}{p}(n-p)$ + $\frac{c}{r}(p-q)$ (? given)

= $\frac{1}{2}[2e+(p-1)d](q-r)$ + $\frac{1}{2}[2e+(q-1)d](r-p)$ + $\frac{1}{2}[2e+(r-1)d](p-q)$

= $\frac{2e}{2}(q-r)+\frac{d}{2}(p-1)(q-r)+\frac{2e}{2}(n-p)$ + $\frac{d}{2}(q-1)(r-p)$ + $\frac{2e(p-q)}{2e}+\frac{d}{2}(r-1)(p-q)$

=c[q – r+r – p+p – q]+ $\frac{d}{2}$ [(p – 1)(q – r)+(q – 1)(r – p)+(r – 1)(p – q)]

=C × 0 + $\frac{d}{2}$ [pq – pr – q+r+qr – pq – r+p+pr=qr – p+q]

=0+ $\frac{d}{2}$ [pq – pq – pr+pr – q+q+r – r+qr+qr+p – p]

= $\frac{d}{2}[0]$

=0.

=R.H.S.

24. Let a and d be the first elements and common different of the A.P.

Then, Sum of first p terms of the AP=Sum of first q terms of A.P

Sp=Sq

$\frac{p}{2}[2a+(p-1)d]=\frac{q}{2}[2a+(q-1)d]$

p [2a+pd – d]=q [2a+qd – d]

2ap+p²d – pd=2aq+q²d – qd

2ap – 2aq+p²d – q²d – pd+qd=0

2a (p – q)+ [p² – q² – p+q]d=0.

2a (p – q)+ [ (p – q) (p+q) – (p – q)]d=0

(p – q) {2a+ [ (p+q) – 1]d}=0

Deviding both sides by p – q,

2a+ [ (p+q) –1]d=0.

And multiplying by P+Q/2 we get,

$\frac{p+q}{2}\{2a+[(p+q)-1]d\}=0$ which in the form $\frac{n}{2}\{2a+(n-1)d\}$ where n=p+q.

S_p+q=0.