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Currently, the admission process for BTech at SASTRA University is closed, which means the university is not accepting any online applications for the BTech course. Usually, the application process starts in the month of April. For the academic year 2025, the application portal opens on Apr 06, 2025. Moreover, the application process concludes in the month of June. In 2025, the application process ends on June 14 2025. Interested applicants should apply for the course in this bracket only. 

Admission to PES University is offered based on merit and entrance. Candidates securing minimum aggregate in Class 12 will be eligible for UG level admissions. Whereas, candidates with clear record of graduation will be eligible for postgraduation admissions. University accepted CAT, MAT, XAT, CLAT, KCET, JEE Main, GATE or other equivalent exam scores for final admission. 

23. Let a1, a2 be the first terms and d1, d2 be the common difference of the first term A.P.S

So, $\frac{\mathrm{Sum\; ofntermofIst}{}^{}AP}{\mathrm{Sum\; of\; n\; terms\; of\; 2nd\; AP}}=\frac{5n+4}{9n+6}$

$\frac{\frac{n}{2}\left[2a_1+(n-1)d_1\right]}{\frac{n}{2}\left[2a_2+(n-1)d_2\right]}=\frac{5n+4}{9n+6}$

$\frac{2a_1+(n-1)d}{2a_2+(n-1)d}=\frac{5n+4}{9n+6}$ ---------------(1)

The ratio of their 18thterms.

$\frac{a_{1\mathrm{8offirstA.P}}}{a_{1\mathrm{8ofsecondA.P}}}=\frac{a_1+(18-1)d_1}{a_2+(18-1)d_2}$

= $\frac{a_1+17d_1}{a_2+17d_2}$

= $\frac{a_1+17d_1}{a_2+17d_2}\times \frac{2}{2}$

= $\frac{2a_1+34d_1}{2a_2+35d_2}$ --------(2)

Comparing eqn (1) and (2),

$\frac{2a_1+34d_1}{2a_2+35d_2}=\frac{2a_1+(35-1)d_1}{2a_2+(35-1)d_1}=\frac{5\times 35+4}{9\times 35+6}$ = $\frac{179}{321}$

So, ratio of their 18th terms = $\frac{179}{321}$

22. Given, sum of n terms of A.P, S_n= (pⁿ+qn²), p&q are constants substituting n=1,

S₁=p× 1+q× 1²=p+q=a₁ [sum upto1^st term only]

And substituting n=2,

S₂=p× 2+q× 2²=2p+4q=a₁+q₂  [sum upto 2^ndtern only]

So,  a₁+a₂=2p+4q

⇒ a₂=2p+4q – a₁=2p+4q – (p+q)=2p – p+4q – q

⇒ a₂=p+3q

So, common difference,  d=a₂ – a₁

= (p+3q) – (p+q)

=p+3q – p – q

=2q.

21. Given, a_k =5k+1

Putting k =1,

a₁ =5 × 1+1=5+1=6.

a_n =5n+1=l

So, sum of unto n teams of the AP, S_n = $\frac{n}{2}[2a+l]$

S_n = $\frac{n}{2}[6+5n+1]$

= $\frac{n}{2}[7+5n]$ .

20. The given A.P. is 25,22,19, …

So, a=25

d=22 – 25= –3

Given that, sum of first n terms of the AP=116.

n [2 × 25+ (n – 1) (–3)]=116 × 2

n [50 – 3n+3]=232

n [53 – 3n]=232 .

53n – 3n²=232.

3n² – 53n+232.

Using quadratic formula, a=3, b= –53, c=232.

n = $\frac{53+5}{6}$ or $\frac{53-5}{6}$

= $\frac{58}{6}$ or $\frac{48}{6}$

= $\frac{29}{3}$ or 8.

As n $\in$ N, n=8.

? Last term=a+ (n – 1)d

=a+ (8 – 1)d

=25+7× (–3)

=25 – 21

=4.

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

Then, given a_p = $\frac{1}{q}$

a+(p – 1)d = $\frac{1}{q}$ --------(1)

and a_q = $\frac{1}{p}$

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

Subtracting eqn (2) from (1) we get,

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

a+(p – 1)d– a–(q– 1)d= $\frac{p-q}{pq}$

[(p – 1)–(q – 1)]d= $\frac{p-q}{pq}$

[p – 1 – q+1]d= $\frac{p-q}{pq}$

[p – q]d= $\frac{p-q}{pq}$ .

d= $\frac{1}{pq}$ .

Putting d= $\frac{1}{pq}$ in eqn (1) we get,

$a+(p-1)\frac{1}{pq}=\frac{1}{q}$ .

$a+\frac{1}{pq}-\frac{1}{pq}=\frac{1}{q}$

$a+\frac{1}{q}-\frac{1}{pq}=\frac{1}{q}$

$a=\frac{1}{q}-\frac{1}{q}+\frac{1}{pq}$ $a=\frac{1}{pq}$ .

So, sum of first pq terms,

$S_{pq}=\frac{pq}{2}\left[2\times \frac{1}{pq}+(pq-1)\frac{1}{pq}\right]$

= $\frac{pq}{2}\times \frac{1}{pq}[2+pq-1]$ .

= $\frac{1}{2}[pq+1]$

PES Bangalore offers courses at the UG, PG and PhD level across the fields of Engineering, Medicine, Architecture, Design, Management and Commerce, Economics and many more, The University offers these courses on the basis of merit and various accepted entrance exams. PES Bangalore fee structure includes a tuition fees ranging from INR 1.9 Lacs to INR 20 Lakh. In comparison to other universities, the cost of academia is higher. However, the University various other facilities apart from academia which is worth the cost being paid. 

18. Let the sum of n farms of the A.P –6, $-\frac{11}{2}$ , –5, …. gives –25.

Then, 

From the given A.P.

a= –6

d= – $\frac{11}{2}-(-6)=-\frac{11}{2}+6=-\frac{11+12}{2}=\frac{1}{2}$

So, –25= $\frac{n}{2}\left[2\times (-6)+(n-1)\left(\frac{1}{2}\right)\right]$

–25 × 2=n $\left[-12+\frac{n-1}{2}\right]$

–50=n $\left[\frac{-12\times 2+(n-1)}{2}\right]$

–50=n $\left[\frac{-24+n-1}{2}\right]$

–50 × 2 =n[n – 25]=n² – 25n

n² – 25n+100=0.

n² – 5n – 20n+100=0

n(n – 5) – 20(n – 5)=0

(n – 5)(n – 20)=0

So, n=5 and n=20.

When n=5

$S_5=\frac{5}{2}\left[-12+(5-1)\frac{1}{2}\right]=\frac{5}{2}\left[-12+4\times \frac{1}{2}\right]=\frac{5}{2}[-12+2]=\frac{5}{2}\times (-10)$ = –25.

When n=20

$S_{20}=\frac{20}{2}\left[-12+(20-1)\frac{1}{2}\right]=10\left[-12+\frac{19}{2}\right]$ = $-120+\frac{19\times 10}{2}$ = –120 + 19 × 5 = –120 + 95 = –25.

17. Given, a=2

Let A1, A2, A3, A4, A5, A6, A1, A6, A7, A10 be the first ten terms. So, A1=a.

Given A1+A2+A3+A4+A5= $\frac{1}{4}$  (A6+A1+A8+A1+A10)

a+ (a+d)+ (a+2d)+ (a+3d)+ (a+4d)

= $\frac{1}{4}$  [ (a+5d)+ (a+6d)+ (a+7d)+ (a+8d)+ (a+9d)]

5a+10d= $\frac{1}{4}$  [5a+35d].

4 [5a+10d]=9a+35d.

20a+40d=5a+35d.

40d – 35d=5a – 20a

5d= –15a

d= –3a

d= –3 * 2 [as a=2]

d= –6

So, A₂₀=a+ (20 – 1)d

=2+19 * (–6)

=2 – 114

= –112.

16. Sum of all natural number between 100 and 1000 which are multiple of 5.

=105+110+115+ … +995.

So, a=105.

a=110 – 105=5.

As the last term is 995 which is the nthterm,

a+ (n – 1)d =995.

105+ (n – 1) × 5 =995.

(n – 1) × 5=995 – 105.

(n – 1)5 =890

n – 1 = $\frac{890}{5}=178$

n =178+1

n =179

So, required sum S_n $\frac{n}{2}$ (a+l); l=last term.

$=\frac{179}{2}(105+995)$

$=\frac{179}{2}\times 1100$

= 179 × 550

=98450.

15. Sum of odd integers from 1 to 2001

=1+3+5+ … +2001

So, a=1

d=3 – 1=2

? For n^th term,

a_n=a+ (n – 1)d.

? the last n^th term is 2001,

2001 =1+ (n – 1)2.

(n – 1)2 =2001 – 1

n – 1= $\frac{2000}{2}=1000$

n =1000+1

n =1001.

? Sum of n terms, S_n= $\frac{n}{2}$  (a + l); l = last term.

? Required sum = $\frac{1001}{2}(1+2001)=\frac{1001}{2}\times 20021001\times 1001$ =1002001

PES University offers BA LLB as a 5 years programme with an annual intake of 60 students. The University offers BA LLB (Hons) on the basis of accepted entrance exam scores of CLAT. Furthermore, PES Bangalore fee structure includes a tuition fees worth INR 11 Lakh. In addition to this, candidates must secure a minimum aggregate in Class 12.

Adverbs are the word that modifies the meaning of a verb, an adjective or another adverb. They tell us about how much, in what manner, how far, in what degree and to what extent. E.g. all, very, probably, very, etc.

• Example: She learns quickly.

Adjectives are the words that add meaning to the nouns or pronouns. They simply make noun and pronoun more descriptive. E.g. beautiful, honest, brave, wealthy.

• Example: She is a quick learner.

As per the Arena Animation, South Extension admission criteria, the college shortlists the candidates based on merit. Interested students for admission needs to score good marks in their academics. There is no need to appear for any entrance exam. 

14. Given, a₁=1=a₂.

a_n=a_{n – 1}+a_{n – 2},n>2.

We need to find, $\frac{a_{n+1}}{n}$

Putting n=3,4,5,6 in an=an – 1+an – 2 we have,

a₃=a_{3 – 1}+a_{3 – 2}=a₂+a₁=1+1=2.

a₄=a_{4 – 1}+a_{4 – 2}=a₃+a₂=2+1=3.

a₅=a_{5 – 1}+a_{5 – 2}=a₄+a₃=3+2=5.

a₆=a_{6 – 1}+a_{6 – 2}=a₅+a₄=5+3=8.

Now, to find $\frac{a_{n+1}}{n}$ ,

Substitute n=1,2,3,4,5.

$\frac{a_{1+1}}{a_1}=\frac{a_2}{a_1}=\frac{1}{1}=1$

$\frac{a_{2+1}}{a_2}=\frac{a_3}{a_2}=\frac{2}{1}=2$

$\frac{a_{3+1}}{a_3}=\frac{a_4}{a_3}=\frac{3}{2}$

$\frac{a_{4+1}}{a_4}=\frac{a_5}{a_4}=\frac{5}{3}$

$\frac{a_{5+1}}{a_5}=\frac{a_6}{a_5}=\frac{8}{5}$ .