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The highest package offered in the R V Institute of Technology and Management placement for BTech CSE in 2025 stood at INR 26.35 LPA. Refer to the table to know the BTech the highest packages offered at R V Institute of Technology and Management placement in 2023-2025:

Note- The placement season is ongoing for 2025, hence the stats may vary once the final report gets released.

86. Given, a = 11

Let d and l be the common difference & last term of the A.P.

Then, $a+(a+d)+(a+2d)+(a+3d)=56$ [first 4 terms sum]

$\Rightarrow 4a+6d=56$

$\Rightarrow 6d=56-4a=56-4\times 11=56-44=12$

$\Rightarrow d=\frac{12}{6}=2$

And, $l+(l-d)+(l-2d)+(l-3d)=112$

$\Rightarrow 4l-6d=112$

$\Rightarrow 4l=112+6d=112+6\times 2=112+12=124$ [last 4 terms sum]

$\Rightarrow l=\frac{124}{4}=31$

So, $l=31$

$\Rightarrow a+{(n-1)}^d=31$

$\Rightarrow 11+{(n-1)}^2=31$

$\Rightarrow (n-1)2=31-11=20$

$\Rightarrow n-1=\frac{20}{2}=10$

$\Rightarrow n=10+1$

$\Rightarrow n=11$

the A.P. has 11 number of terms.

There are a total of 30 seats for BTech in Computer Science and Engineering. Candidates are admitted as per the sanctioned seat intake and performance in the selection rounds. It is within the rights of the university to revise its seat count. Moreover, the mentioned seat count is as per the official website/ sanctioning body. It is still subject to revision and hence, is indicative. 

Yes, at Somaiya School of Design, supported by Riidl, students officially enrolled in the BDes course are offered scholarships. The institute offers various scholarships. Listed below are some of the scholarships that can be availed by the BDes students:

• EWS (Economically Weaker Section) Scholarship
• For Students with Disabilities (PWD)
• For Wards of Defense Personnel

85. Let a and r be the first term and common ratio of G.P.

Then, number of term = 2n (even).

$a_1+a_2+?+a_{2n}=5\left(a_1+a_3+?+a_{2n-1}\right)[?]$

$\Rightarrow a+ar+.........+a{r}^{2n-1}=5\left(a+a{r}^2+.......+a{r}^{2n-1-1}\right)$

$\Rightarrow \frac{a\left(1-r^{2n}\right)}{1-n}=5\times \frac{a\left[1-{\left(r^2\right)}^n\right]}{1-r^2}$ { series on R.H.S. has $\frac{2n}{2}=n$ terms and common ratio $\frac{a{r}^2}{a}=r^2$ }

$\Rightarrow \frac{1-r^{2n}}{1-r}=\frac{5\left[1-r^{2n}\right]}{1-r^2}$ (eliminating a)

$\Rightarrow \frac{1-r^{2r}}{1-r}=\left[\frac{1-r^{2r}}{1-r}\right]\times \frac{5}{1+r}\left\{?a^2-b^2=(a-b)(a+b)\right\}$ 

$1=\frac{5}{1+r}\{Eliminating\; same\; term\}$

$\Rightarrow 1+r=5$

$\Rightarrow r=5-1$

r = 4

84. Let a, ar and $a{r}^2$ be the three nos. which is in G.P.

Then, a + ar + ar² = 56

a ( 1 + r + r²) =56  -I

Given, that a1, ar 7, ar² - 21 from an AP we have,

$(ar-7)-(a-1)=\left(a{r}^2-21\right)-(ar-7)$

$\Rightarrow ar-7-a+1=a{r}^2-21-ar+7$

$\Rightarrow ar-a-6=a{r}^2-ar-14$

$\Rightarrow a{r}^2-ar-ar+a-a-14-6$

$\Rightarrow a{r}^2-2ar+a=8$

$\Rightarrow a\left(r^2-2r+1\right)=8$ ………………. II

Now, dividing equation I by II we get,

$\Rightarrow \frac{a\left(1+n+n^2\right)}{a\left(r^2-2r+1\right)}=\frac{56}{8}$

$\Rightarrow 1+r+r^2=7\left(n^2-2n+1\right)$

$\Rightarrow 1+r+r^2=7r^2-14n+7$

$\Rightarrow 7r^2-14n+7-1-x-r^2=0$

$\Rightarrow 6r^2-15r+6=0$

$\Rightarrow 2r^2-5r+2=0$ (dividing by 3 throughout)

$\Rightarrow 2r^2-4r-r+2=0$

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

$\Rightarrow (r-2)(2r-1)=0$

$\Rightarrow r-2=02r-1=0$

$\Rightarrow r=2r=\frac{1}{2}$

So, when r = 2, putting in equation I,

$\Rightarrow a\left(1+2+2^2\right)=56$

$\Rightarrow a(1+2+4)=56$

$\Rightarrow a(7)=56$

$\Rightarrow a=\frac{56}{7}=8$

The numbers are 8, 8× 2, 8× 2² = 8, 16, 32.

And When $r=\frac{1}{2}$ putting in equation I,

$a\left(1+\frac{1}{2}+\frac{1}{2^2}\right)=56$

$\Rightarrow a\left(1+\frac{1}{2}+\frac{1}{4}\right)=56$

$\Rightarrow a\left(\frac{4+2+1}{4}\right)=56$

$\Rightarrow a\times \frac{7}{4}=56$

$\Rightarrow a=56\times \frac{4}{7}=32$

So, the numbers are $32,32\times \frac{1}{2},32\times {\left(\frac{1}{2}\right)}^2\Rightarrow 32,16,8$

The average package offered at R V Institute of Technology and Management placements stood at INR 11.47 LPA to the BTech graduates. Check out the table for reference:

For admission to Somaiya School of Design, supported by Riidl BDes course, students can apply with valid JEE Main Paper 2 scores. For the 2025 session, the test had already been conducted on Apr 9, 2025. Moreover, the exam results are expected to be released in the fourth week of May, 2025.

83. Given, a = 1

$a_3+a_5=90$

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

So,

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

So,

$a_3+a_5=a{r}^{3?1}+a{r}^{5?1}=90$

$?a\left[r^2+r^4\right]=90$

$?1\left[r^2+r^4\right]=90\text{?}\{?a=1\}$

$?r^4+r^2?90=0$

Let $x=r^2$ so we can write above equation as

$x^2+x?90=0x=r^2$

82. Given, a = 5

$r=2>1$

$s_n=315$

So,  $\frac{a\left(r^n-1\right)}{r-1}=315$

$\Rightarrow \frac{5\left(2^n-1\right)}{2-1}=315$

$\Rightarrow 2^n-1=\frac{315}{5}$

$\Rightarrow 2^n=63+1$

$\Rightarrow 2^n=64=26$

$\therefore n=6$

Hence, last term $=a_6=a{r}^{6-1}=a{r}^5=5\times 2^5=5\times 32$

= 160

81. Given, $f(x+y)=f(x)\cdot f(y).Ux,y\in N$ and $f\left(1\right)=3$

Putting (x, y) = (1+1) we get

Putting (x, y) = (1,1) we get,

$f(1+1)=f(1)\cdot f(1)=3\cdot 3=9$

$\Rightarrow f(2)=9$

And putting $(x,y)=(1,2)$ we get,

$f(1+2)=f(1)\cdot f(2)=3\times 9=27$

$f(3)=27$

$f(1)+f(2)+f(3)+?+f(x)={\displaystyle \sum _{x=1}^{n}f}(x)=120$ (Given)

As, With a = 3

$r=\frac{9}{3}=3>1$

We can write equation I as ,

$\frac{a\left(r^n-1\right)}{r-1}=120$

$\Rightarrow \frac{3\left(3^n-1\right)}{3-1}=120$

$\frac{3}{2}\left(3^n-1\right)=120$

$\Rightarrow \left(3^n-1\right)=120\times \frac{2}{3}$

3ⁿ 1 = 80

3ⁿ = 81 +1

3ⁿ = 81

3ⁿ = 3⁴

n = 4

The Somaiya School of Design, supported by Riidl BDes course fees is calculated by adding multiple fee elements essential to be paid by the students to complete the course. As per the fee structure, students must pay two major components: tuition fees and a one-time fee. Upon adding all, the total fees for the BDes programme amount to INR 21.29 lakh. Check the below table to learn the component-wise bifurcation of the total fees:

Note: The above-mentioned fee is as per the official sources. However, it is indicative and subject to change.

80. Two digits no. when divided by 4 yields 1 as remainder are, 12+1, 16+1, 20+1 …., 96+1

13, 17, 21, ………97 which forms an A.P.

So, a = 13

$d=17-13=4$

$l=97$

$\Rightarrow a+(x-1)d=97$

$\Rightarrow 13+(x-1)4=97$

$\Rightarrow \left(x-1\right)4=97-13=84$

$\Rightarrow x-1=\frac{84}{4}=21$

$\Rightarrow x=21+1=22$

Sum of numbers in A.P. = $\frac{x}{2}(a+l)$

$=\frac{22}{2}(13+97)$

= 11× 110

= 1210

R V Institute of Technology and Management provides good placements to its graduating BTech 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 R V Institute of Technology and Management placement in 2023-2024:

79. An A.P. of numbers from 1 to 100 divisible by 2 is

2, 4, 6, ……….98, 100.

So, a = 2 and d = 4-2 = 2

$l=$ 100

$\Rightarrow a+(n-1)d=100$

$\Rightarrow 2+(n-1)2=100$

$\Rightarrow \frac{2}{2}+(n-1)\frac{2}{2}=\frac{100}{2}$

$\Rightarrow 1+(n-1)=50$

$\Rightarrow n=50$

Let, $s_1=2+4+6?+9+100=\frac{50}{2}[2+100]\left\{?S_n=\frac{n}{2}(a+1)\right\}$

$=\frac{50\times 102}{2}$

= 2550

Similarly, an A.P. of numbers from 1 to 100 divisible by 5 is

5,10,15, ……. 95,100

So, a = 5 and d = 100

$l=$ 100

$\Rightarrow a+(n-1)d=100$

$\Rightarrow 5+(n-1)5=100$

$\Rightarrow 1+(n-1)=20$

$\Rightarrow n=20$

$\therefore S_2=5+10+15\dots +95+100=\frac{20}{2}[5+100]$

$=\frac{20\times 105}{2}$

= 1050

As there are also no divisible by both 2 and 5 , i.e., LCM of 2 and 5 = 10 An A.P. of no. from 1 to 100 divisible by 10 is 10, 20, …………100

So, a = 10, d = 10

$l=100$

$\Rightarrow a+(x-1)l=100$

$\Rightarrow 10+(x-1)10=100$

$\Rightarrow 1+(x-1)=10$

$\Rightarrow x=10$

So, $S_3=10+20+30+\dots +100=\frac{10}{2}[10+100]$

= 550

The required sum of number $=S_1+S_2{S}_3=2550+1050-550=3050$

= 3050

78. $\begin{array}{ccccc}\hfill 7\frac{28}{200}\hfill & \hfill \frac{57}{400}\hfill & \hfill \frac{14}{\frac{60}{\frac{56}{4}}}\hfill & \hfill \frac{35}{\begin{array}{l}50\\ \frac{49}{1}\\ \end{array}}\hfill & \hfill \hfill \\ \hfill \hfill & \hfill Smallest =200-4+7=203\hfill & \hfill \hfill & \hfill Largest =400-1=399\hfill & \hfill \hfill \end{array}$

So we can form an A.P. 203,203+7, …., 399-7, 399

So, i.e. a = 203 and d = 7 $$

l = 399

$\Rightarrow a+(x-1)d=399$

$\Rightarrow 203+(x-1)7=399$

$\Rightarrow (x-1)7=399-203$

$\Rightarrow x-1=\frac{196}{7}$

$\Rightarrow x=28+1$

$\Rightarrow x=29$

Sum of the 29 number of the AP = Required sum = $\frac{\eta }{2}[a+l)$

$=\frac{29}{2}[203+399]$

$=\frac{29}{2}\times 602$

= 8729.

BTech admission process at NIILM University is merit-based. Selection is done by evaluating the academic performance of students in their Class 12 examinations (PCM stream) or based on a Diploma in engineering for lateral entry.

Yes, in case a student wishes to withdraw BDes admission from Somaiya School of Design, supported by Riidl, he/she is entitled to a fee refund. The institute is affiliated with Somaiya Vidyavihar University and thus follows the refund policy set by it. Check out the below table to know the amount of refund: