(i) (a): A.P
200, 350, 500... It forms an A.P.
(ii) (b): 950
We know that,Tn = a+ (n– 1)
Here, a = 200, d = 150
ஃ T⁶ = 200 + 5 × 150
= 200 + 750
= 950
(iii) (d): 8750
Using formula,\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\) 
we get, \(S_{10}=\frac{10}{2}[400+9 \times 150]\) 
= 8750
He pays Rs. 8750 in 10 months.
(iv) (a): 150 n + 50
Using formula, T_n = a + (n – 1) d
we get a_n = 200 + (n – 1) × 150
= 200 – 150 + 150 n
= 50 + 150 n
(v) (a): 20
\(S_{n}=\frac{n}{2}[2 \times 200+(n+1) \times 150]\) 
\(\Rightarrow\ 32500=\frac{n}{2}[250 n+50]\) 
⇒ 5n²+ 150n= 32500
⇒ 3n² + 5 n – 1300 = 0
⇒ (n – 20)(3n + 65) = 0
⇒ n = 20 \(\left[\because n \neq \frac{-65}{3}\right]\) 

(i) (a): 25
Explanation: Here, a₃ = a +(3 – 1) d
= a + 2d
= 600 ........(i)
a₇ = a + (7 – 1)
= 700 ........(ii)
Solving (i) and (ii), we get
d = 25
(ii) (c): 550
Substituting the value of d in (i), we get a = 550
(iii) (d): 4800
We know that, \(S_{n}=\frac{n}{2}[2 a+(n-1) d]\) 
Here, n = 8,a= 550 and d = 25
\(S_{8}=\frac{8}{2}(2 \times 550+(4) 25)\)
 S₈ = 4(1100 +100)
= 4800
(iv) (c): a_n–a_n–1 
Difference between the nth and (n– 1)th term is common difference.
(v) (a): 2450
\(S_{4}=\frac{4}{2}[2 \times 550+(3) 25]\) 
= 2(1100+75)
= 2350
and S₈ = 4800
Therefore, difference = 4800 – 2350
= 2450

(i) (d): 2860
Saurabh can choose four cards from same suit in 4 x ¹³ C₄ way
\( =4 \times \frac{13 !}{9 ! \times 4 !} \)
\(=4715=2860\)
(ii) (b): 28561
Here one card to be selected from each suit therefore, he can select in ¹³C₁ x ¹³C₁ x ¹³C₁ x ¹³C₁ ways
= (¹³C₁)⁴ = 28561
(iii) (d): 495
There are 12 face cards and 4 are to be selected out of these 12 cards. This can be done in ¹² C₄ ways
\(=\frac{12 !}{8 ! 4 !} 495\)
(iv) (c): 105625
Two red and two black cards can be selected in ²⁶C₂ \(\times{ }^{26} C_{2} \text { ways }=\left(\frac{26 !}{24 ! 2 !}\right)^{2} 105625\)
(v) (d): 650
Manoj can select two cards of same colour in ²⁶ C₂ + ²⁶C₂ ways = 325 + 325 = 650

(i) (d): 120
Total number of ways = 5! = 120
(ii) (c): 48
Two position are fixed for Rahul and Chetan therefore considering it as one unit, total students left = 3 + 1 = 4
Total possible arrangement = 4! x 2! = 48
(iii) (a): 12
Total possible arrangements = 3! x 2! = 12
(iv) (d): 24
Total possible arrangements = 4! = 24
(v) (b): 12
Total possible arrangements = 2! x 3! = 12

(i) (b): 5
We can make the following Venn diagram to understand the problem easily:

Here, we have: n(C∩M∩P) = 3 (given), where Crepresents Chemistry, P represents Physics and Mrepresents Mathematics.
n(U) = Total students = 25
By using Venn diagram, we can easily see thatstudents who read only chemistry are 5
(ii) (c): 11
Here we have to find n(C) + n(P) + n(M) = 5 + 2 + 4 = 11
Therefore, total students who read only one subject
= 11
(iii) (a): 23
Here we need to find:
n(P∪C∪M) = n(M) +n(P) + n(C) – n(P∩C)–n(P∩M) –n(M∩C) + n(P∩C∩M)
= 23
(adding all the numbers from venn diagram will give us result).
(iv) (d): 2
Here we need to findn(M∪C∪P)'
= n(U) – n(P∪C∪M)
= 25 – 23 = 2
(v) (c): 70
Since total student who read only Chemistry book are 5, therefore total time devoted to read only chemistry book by these students are: 514 = 70 hrs.

(i) (d): 20
Given that, n(T) = 12, n(C) = 15 n(T∩C) = 7
As we know that:
n (T∪C) = n (T) + n(C) – n(T∩C)
= 12 + 15 – 7 = 20
Therefore total person who like atleast one of the two drinks were 20.
(ii) (c): 5
To find total number of person who like only tea but not coffee we can use :
= n(T) – n(T∩C)
= 12 – 7 = 5
(iii) (d): 8
Only coffee but not tea
= n(C) – n(T∩C)
= 15 – 7
= 8
(iv) (c): 
Neither tea nor coffee
= n(U) – n(T∪C)
= 25 – 20
= 5
(v) (d): 1550
Total number of person like only Tea = 5 and total number of person who liked only Coffee = 8
So, total cost will be: 5 x 2 x 35 + 8 x 2 x 75
[ As each person had taken two cup of tea and coffee ]
= 350 + 1200 = 1550
So, total cost for tea and coffee will be RS. 1550