Class 11 Maths, Sets and Sequence and Series Test 1 and Solutions – 2027

This is the solution for the test Class 11 Sets and Sequence and Series Test – 1. Let’s break down each problem step-by-step. Designed for both students and teachers, these answers make exam revision easier with clear explanations, formula highlights, and downloadable PDF options.

Section A: 1 Mark Questions

Q1. Two finite sets have m and n elements. The number of subsets of the first set is 112 more than that of second set. The values of m and n are respectively:

(a) 4, 7
(b) 7, 4
(c) 4, 4
(d) 7, 7

Solution:
Step 1: Recall the formula for the number of subsets.
If a finite set contains k elements, the total number of subsets it can form is exactly 2^k.

Step 2: Set up the mathematical equation.
Number of elements in the first set = m
Number of subsets of the first set = 2^m

Number of elements in the second set = n
Number of subsets of the second set = 2^n

The problem clearly states that the first set has 112 more subsets than the second set.
We can write this relationship as:
2^m - 2^n = 112

Step 3: Analyze and factor the equation.
Since 112 is a positive integer, it is obvious that 2^m must be greater than 2^n, which tells us that m > n.

To solve this algebraically without guessing, we take out the smaller term (2^n) as a common factor from the left-hand side:
2^n(2^{m-n} - 1) = 112

Step 4: Prime factorize the numerical value.
Now, we need to break down 112 into a product of an even number (a power of 2) and an odd number, because our left-hand side is structured exactly that way (2^n is even, and 2^{m-n} - 1 is always odd).

Let’s factorize 112:
112 = 16 \times 7
112 = 2^4 \times 7

So, our equation perfectly balances out as:
2^n(2^{m-n} - 1) = 2^4 \times 7

Step 5: Compare both sides.
By comparing the even parts (the powers of 2) and the odd parts on both sides of the equation, we get two very simple relations:

2^n = 2^4
2^{m-n} - 1 = 7

Step 6: Solve for m and n.

From our first relation, the bases are the same, so we directly equate the powers:
n = 4

Now, substitute n = 4 into our second relation:
2^{m-n} = 7 + 1
2^{m-n} = 8
2^{m-4} = 2^3

Again, the bases are the same, so we equate the powers:
m - 4 = 3
m = 7

Final Answer: \boxed{ m = 7, \ \ n = 4 }
Correct option: (b).

Q2. If the third term of G.P. is 4, then the product of its first 5 terms is:

(a) 4^3
(b) 4^4
(c) 4^5
(d) none of these

Solution:
Step 1: Define the terms of the Geometric Progression (G.P.).
Let the first term of the G.P. be a and the common ratio be r.
The n-th term of a G.P. is given by the formula a_n = a r^{n-1}.

Step 2: Use the given information to create an equation.
We are given that the third term is 4.
So, a_3 = a r^{3-1} = a r^2 Therefore, a r^2 = 4.

Step 3: Write out the product of the first 5 terms.
The first 5 terms of the G.P. are: a, a r, a r^2, a r^3, a r^4.

Their product is:
P = a \times a r \times a r^2 \times a r^3 \times a r^4

Step 4: Simplify the product.
Multiply the terms together by adding the exponents of a and r:
P = a^5 r^{1+2+3+4} = a^5 r^{10}

Step 5: Relate the product back to our known value.
We can rewrite a^5 r^{10} using the laws of exponents as (a r^2)^5.

Since we found in Step 2 that a r^2 = 4,
we substitute this value into our product:
P = (4)^5 = 4^5

Final Answer: \boxed{ 4^5 }
Correct option: (c)

Teacher’s Tip: In G.P. product questions involving an odd number of terms (like 5), the product is always the middle term raised to the power of the number of terms. Here, the middle term is the 3rd term, so the product is simply (3\text{rd term})^5!

Q3. Let A and B be two sets such that n(A)=20, n(B)=10, n(A\cup B)=15. Then, n(A\cap B) is equal to:

(a) 30
(b) 40
(c) 15
(d) none of these

Solution:
Step 1: Recall the Principle of Inclusion-Exclusion for two sets.
The formula relating the number of elements in the union and intersection of two sets is:
n(A \cup B) = n(A) + n(B) - n(A \cap B)

Step 2: Substitute the given values into the formula.
n(A) = 20
n(B) = 10
n(A \cup B) = 15

Substitute these into our equation:
15 = 20 + 10 - n(A \cap B)

Step 3: Solve for n(A \cap B).
15 = 30 - n(A \cap B)

Rearrange the equation to isolate n(A \cap B):
n(A \cap B) = 30 - 15
n(A \cap B) = 15

Final Answer: \boxed{ 15 }
Correct option: (c)

Q4. If A=\{x:x \text{ is a multiple of 3}\} and, B=\{x:x \text{ is a multiple of 5}\}, then A\cap B is:

(a) \{x: x \text{ is a multiple of 3}\}
(b) \{x: x \text{ is a multiple of 5}\}
(c) \{x: x \text{ is a multiple of 15}\}
(d) none of the above

Solution:
Step 1: Understand the definition of sets A and B.
Set A contains all multiples of 3: A = \{3, 6, 9, 12, 15, 18, 21, 24, 27, 30, \dots\}
Set B contains all multiples of 5: B = \{5, 10, 15, 20, 25, 30, 35, \dots\}

Step 2: Find the intersection A \cap B.
The intersection of two sets contains the elements that are common to both.
We need numbers that are multiples of both 3 and 5.
Looking at our lists, the common elements are 15, 30, 45, \dots

Step 3: Generalize the pattern.
A number is a multiple of both 3 and 5 if and only if it is a multiple of their Least Common Multiple (LCM).
\text{LCM}(3, 5) = 15.
Therefore, the common elements are all multiples of 15.

Step 4: Write the set in set-builder notation.
A \cap B = \{x: x \text{ is a multiple of 15}\}

Final Answer: \boxed{ \{x: x \text{ is a multiple of 15}\} }
Correct option: (c)

Q5. 18th term from the end of the sequence 3, 6, 12,…25th term is:

(a) 393216
(b) 393206
(c) 384
(d) 768

Solution:
Step 1: Identify the type of sequence and its parameters.
The sequence is 3, 6, 12, … Since 6/3 = 2 and 12/6 = 2, this is a Geometric Progression (G.P.). First term, a = 3.
Common ratio, r = 2.
The question states the sequence goes up to the 25th term, so total terms n = 25.

Step 2: Address the wording of the question.
The question asks for the “18th term from the end”, which mathematically corresponds to position 25 - 18 + 1 = 8 from the beginning.

Step 3: Calculate the 18th term from the beginning.
The formula for the m-th term of a G.P. is a_m = a r^{m-1}.

Here, m = 8.

a_{8} = 3 \times 2^{8-1}

a_{8} = 3 \times 2^{7}

Step 4: Compute the final value.
We know that 2^7 = 128.

a_{18} = 3 \times 128 = 384.

Final Answer: \boxed{ 384 }
Correct option: (c)

Q6. If X=\{8^n-7n-1\vert{}n\in N\} and Y=\{49n-49\vert{}n\in N\}. Then

(a) X\subset Y
(b) Y\subset X
(c) X=Y
(d) X\cap Y=\phi

Solution:
Step 1: Analyze set X using the Binomial Theorem.
The elements of X are defined by the expression 8^n - 7n - 1.
We can rewrite the base 8 as (1 + 7)^n.

Using the Binomial expansion: (1 + 7)^n = 1 + \binom{n}{1}7^1 + \binom{n}{2}7^2 + \binom{n}{3}7^3 + \dots + 7^n
(1 + 7)^n = 1 + 7n + \binom{n}{2}7^2 + \binom{n}{3}7^3 + \dots + 7^n

Step 2: Substitute this back into the expression for X.
8^n - 7n - 1 = (1 + 7n + \binom{n}{2}7^2 + \dots + 7^n) - 7n - 1

The 1 and 7n terms neatly cancel out:
8^n - 7n - 1 = \binom{n}{2}7^2 + \binom{n}{3}7^3 + \dots + 7^n

Step 3: Factor out the common term.
We can factor out 7^2 (which is 49) from the remaining terms:
8^n - 7n - 1 = 49 \left[ \binom{n}{2} + \binom{n}{3}7 + \dots + 7^{n-2} \right]

This proves that for any natural number n, the value is a multiple of 49.
Plugging in n=1, 2, 3..., we get X = \{0, 49, 490, \dots\}.

Step 4: Analyze set Y.
The elements of Y are defined by 49n - 49, which factors to 49(n - 1).
For n \in N (i.e., n = 1, 2, 3, \dots), the generated values are:
Y = \{0, 49, 98, 147, \dots\}.
Set Y contains all non-negative multiples of 49.

Step 5: Compare the two sets.
Set Y contains every single multiple of 49.
Set X contains only some multiples of 49 (skipping many, like 98 and 147).

Therefore, every element present in X is also found in Y, meaning X is a proper subset of Y.

Final Answer: \boxed{ X \subset Y }
Correct option: (a)

Section B: 2 Marks Questions

Q7. If X=\{5,6,7,8\}, Y=\{7,8,9,10\}, Z=\{3,4,5,6\}. Find:
(a) (X\cap Y)\cup Z
(b) (X\cup Y) \cap Z

Solution:
Step 1: Find the intersection and union for the inner brackets.
(a), first find X \cap Y (elements common to X and Y):
Given X=\{5,6,7,8\} and Y=\{7,8,9,10\}, their common elements give X \cap Y = \{7, 8\}.

We need (X \cap Y) \cup Z.
Take the set we found, \{7, 8\}, and combine it with Z = \{3, 4, 5, 6\}.
(X \cap Y) \cup Z = \{3, 4, 5, 6, 7, 8\}

(b), first find X \cup Y (all elements in X and Y combined, without duplicates):
X \cup Y = \{5, 6, 7, 8, 9, 10\}.

We need (X \cup Y) \cap Z.
Take the set X \cup Y = \{5, 6, 7, 8, 9, 10\} and find common elements present in Z = \{3, 4, 5, 6\}.
(X \cup Y) \cap Z = \{5, 6\}

Final Answer: \boxed{ \text{(a) } \{3, 4, 5, 6, 7, 8\}, \text{ (b) } \{5, 6\} }

Q8. Insert five numbers between 8 and 26 such that resulting sequence is an A.P.

Solution:
Step 1: Define the Arithmetic Progression (A.P.).
Let the numbers be A_1, A_2, A_3, A_4, A_5.

The sequence becomes: 8, A_1, A_2, A_3, A_4, A_5, 26.
This forms an A.P. with a total of n = 7 terms.
First term, a = 8. Last term, a_7 = 26

Step 2: Use the A.P. formula to find the common difference (d).
The formula for the n-th term is a_n = a + (n-1)d.
a_7 = a + (7-1)d
26 = 8 + 6d

Step 3: Solve for d.
26 - 8 = 6d
18 = 6d
d = 3

Step 4: Find the five inserted numbers.
Add the common difference d=3 sequentially starting from the first term.
A_1 = 8 + 3 = 11
A_2 = 11 + 3 = 14
A_3 = 14 + 3 = 17
A_4 = 17 + 3 = 20
A_5 = 20 + 3 = 23

Final Answer: \boxed{ 11, 14, 17, 20, 23 }

Teacher’s Tip: To quickly verify your work in the exam hall, just add the common difference to your last inserted number.
23 + 3 = 26, which perfectly matches our required final term!

Q9. Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Solution:
Step 1: Understand the task using Set Theory logic.
Let S_2 be the sum of integers divisible by 2.
Let S_5 be the sum of integers divisible by 5.
Let S_{10} be the sum of integers divisible by both 2 and 5 (i.e., multiples of 10).

To avoid double-counting the numbers divisible by both, the required total sum is
S = S_2 + S_5 - S_{10}.

Step 2: Calculate S_2 (Sum of multiples of 2).

Multiples of 2 up to 100: 2, 4, 6, \dots, 100.
A.P. with a=2, d=2, and last term l=100.
Number of terms n = 100/2 = 50.

Using the sum formula S_n = \frac{n}{2}(a + l):

S_2 = \frac{50}{2}(2 + 100) = 25 \times 102 = 2550.

Step 3: Calculate S_5 (Sum of multiples of 5).

Multiples of 5 up to 100: 5, 10, 15, \dots, 100.
A.P. with a=5, d=5, l=100.
Number of terms n = 100/5 = 20.

S_5 = \frac{20}{2}(5 + 100) = 10 \times 105 = 1050.

Step 4: Calculate S_{10} (Sum of multiples of 10).
Multiples of 10 up to 100: [/latex]10, 20, 30, \dots, 100[/latex].
A.P. with a=10, d=10, l=100.
Number of terms n = 100/10 = 10.

S_{10} = \frac{10}{2}(10 + 100) = 5 \times 110 = 550.

Step 5: Calculate the final sum.
S = S_2 + S_5 - S_{10}
S = 2550 + 1050 - 550 = 3050.

Final Answer: \boxed{ 3050 }

Section C: 3 Marks Questions

Q10. Let A=\{1,2,3,4\}, B=\{1,2,3\} and C=\{2,4\}. Find all sets of X satisfying each pair of conditions:
(a) X\subset B and X \not\subset C
(b) X\subset B, X\ne B and X \not\subset C
(c) X\subset A, X\subset B and X\subset C.

Solution:
List all possible subsets of B.
Since B = \{1, 2, 3\},
its subsets are:
\phi, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}

(a).
Condition 1: X \subset B.
This means X must be one of the 8 sets listed in Step 1.

Condition 2: X \not\subset C.
The set C = {2, 4}.
We must eliminate any set from our list that is a subset of C.
The subsets of B that are also subsets of C are \phi and {2}.

Removing these gives:
X = \{1\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}

(b).
This adds the condition X \ne B.
We take our answer from part (a) and simply remove the set B itself ({1, 2, 3}).

X = \{1\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}

(c).
Condition: X \subset A, X \subset B, and X \subset C.
This means X must be a subset of the intersection of all three sets: A \cap B \cap C.
A \cap B \cap C = \{1, 2, 3, 4\} \cap \{1, 2, 3\} \cap \{2, 4\} = \{2\}
The subsets of \{2\} are \phi and \{2\}.

Final Answer:
(a) \boxed{ \{1\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\} }

(b) \boxed{ \{1\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\} }

(c) \boxed{ \phi, \{2\} }

Q11. Let P and Q be sets, if P\cap X=Q\cap X=\phi and P\cup X=Q\cup X for some set X. Show that P=Q.

Solution:
Step 1: Express set P using the intersection property.
We know from set theory that any set P intersected with a superset of itself equals P.
Therefore, P = P \cap (P \cup X).

Step 2: Substitute the given union condition.
We are given that P \cup X = Q \cup X.
Substitute this into our equation from Step 1: P = P \cap (Q \cup X)

Step 3: Apply the Distributive Law of sets.
P = (P \cap Q) \cup (P \cap X)

Step 4: Apply the given intersection condition.
We are given that P \cap X = \phi.
Substitute this into our equation:
P = (P \cap Q) \cup \phi

P = P \cap Q

This mathematically proves that P is a subset of Q (i.e., P \subset Q).

Step 5: Repeat the exact logic for set Q.
Q = Q \cap (Q \cup X)

Q = Q \cap (P \cup X) (since Q \cup X = P \cup X)

Q = (Q \cap P) \cup (Q \cap X)

Q = (Q \cap P) \cup \phi (since Q \cap X = \phi)

Q = P \cap Q

This proves that Q is a subset of P (i.e., Q \subset P).

Final Answer: Since P \subset Q and Q \subset P, it is proved that \boxed{ P = Q } .

Teacher’s Tip: Proving two sets are equal almost always involves proving they are subsets of each other. The identity A = A \cap (A \cup B) is a powerful trick to start proofs like this!

Q12.a If the pth, qth and rth term of a G.P. are a, b, c respectively, prove that: a^{q-r} b^{r-p} c^{p-q} = 1.

Solution:
Step 1: Define the G.P.
Let the first term of the G.P. be A and the common ratio be R.

Step 2: Write the terms mathematically.
a = A R^{p-1}

b = A R^{q-1}

c = A R^{r-1}

Step 3: Substitute into the LHS of the expression to be proved.
LHS = a^{q-r} \cdot b^{r-p} \cdot c^{p-q}

LHS = (A R^{p-1})^{q-r} \cdot (A R^{q-1})^{r-p} \cdot (A R^{r-1})^{p-q}

Step 4: Group the bases A and R.
LHS = A^{(q-r) + (r-p) + (p-q)} \cdot R^{(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)}

Step 5: Simplify the exponents.
Power of A = q - r + r - p + p - q = 0

Power of R = (pq - pr - q + r) + (qr - qp - r + p) + (rp - rq - p + q) = 0

LHS = A^0 \cdot R^0 = 1 \cdot 1 = 1 = RHS. Hence proved.

OR

Q12.b The 5th, 8th and 11th term of a GP are p, q and s, respectively. Show that q^2=ps.

Solution:
Step 1: Define the G.P.
Let the first term be a and the common ratio be r.

Step 2: Write the 5th, 8th, and 11th terms.
5th term: p = a r^4
8th term: q = a r^7
11th term: s = a r^{10}

Step 3: Calculate q^2 and ps.
LHS: q^2 = (a r^7)^2 = a^2 r^{14}

RHS: ps = (a r^4)(a r^{10}) = a^2 r^{4+10} = a^2 r^{14}

Final Answer: LHS = RHS, hence \boxed{ q^2 = ps } is proved.

Q13. If A>0, B>0 and A+B=\pi/6, then what is the minimum value of \tan A + \tan B?

Solution:

Step 1: Simplify the expression using sine and cosine.

\tan A + \tan B = \frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}

Taking a common denominator:

= \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B}

Step 2: Apply the compound angle formula for sine.

The numerator is the expansion of \sin(A+B).

= \frac{\sin(A+B)}{\cos A \cos B}

Step 3: Substitute the given value of A+B. We know A+B = \pi/6 = 30^\circ. Thus, \sin(\pi/6) = 1/2. Expression = \frac{1/2}{\cos A \cos B}

Step 4: Maximize the denominator to minimize the overall value.

To find the minimum value of the fraction, we must find the maximum value of \cos A \cos B.

Multiply numerator and denominator by 2:

= \frac{1}{2 \cos A \cos B}

Using the product-to-sum formula: 2 \cos A \cos B = \cos(A+B) + \cos(A-B)

Denominator = \cos(\pi/6) + \cos(A-B) = \frac{\sqrt{3}}{2} + \cos(A-B)

Step 5: Find the maximum of \cos(A-B).

The maximum value of cosine is 1, which happens when the angle is 0.

So, A - B = 0 \Rightarrow A = B.

Since A+B = 30^\circ, this means A = 15^\circ and B = 15^\circ.

Max denominator = \frac{\sqrt{3}}{2} + 1 = \frac{\sqrt{3} + 2}{2}

Step 6: Calculate the final minimum value.

Minimum Value = \frac{1}{\frac{\sqrt{3} + 2}{2}} = \frac{2}{2 + \sqrt{3}}

Rationalizing the denominator:

= \frac{2(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{2(2 - \sqrt{3})}{4 - 3} = 4 - 2\sqrt{3}

Final Answer: \boxed{ 4 - 2\sqrt{3} }

Teacher’s Tip: You can also solve this instantly using Jensen’s Inequality! The function f(x) = \tan x is strictly convex in the first quadrant. Therefore, the minimum occurs exactly when the variables are equal, meaning A = B = 15^\circ. Just calculate 2 \tan(15^\circ)!

Q13. The sum of first three terms of a Geometric progression is 13/12 and their product is -1. Find the common ratio and the terms.

Solution:
Step 1: Choose the terms of the G.P. wisely.
Whenever a question gives you the product of an odd number of terms in a Geometric Progression, never take the terms as a, ar, ar^2.
Instead, take the middle term as a and divide/multiply by the common ratio r.
Let the first three terms of the G.P. be: \frac{a}{r}, \ a, \ ar

Step 2: Apply the product condition to find ‘a‘.
We are given that the product of these three terms is -1.
\left(\frac{a}{r} \right) \times (a) \times (ar) = -1

a^3 = -1

Since we are dealing with real numbers, taking the cube root gives:
a = -1

Step 3: Apply the sum condition to form an equation.
We are given that the sum of the three terms is 13/12.

\frac{a}{r} + a + ar = \frac{13}{12}

Now, substitute the value of a = -1 that we found in Step 2:
\frac{-1}{r} - 1 - r = \frac{13}{12}

Step 4: Simplify and form a quadratic equation.
Take the negative sign common on the left side:
-\left(\frac{1}{r} + 1 + r \right) = \frac{13}{12}

-\left(\frac{1 + r + r^2}{r} \right) = \frac{13}{12}

Now, cross-multiply to get rid of the denominators:
-12(1 + r + r^2) = 13r

-12 - 12r - 12r^2 = 13r

0 = 12r^2 + 25r + 12

12r^2 + 25r + 12 = 0

Step 5: Solve the quadratic equation by splitting the middle term.
12r^2 + 16r + 9r + 12 = 0

4r(3r + 4) + 3(3r + 4) = 0

(4r + 3)(3r + 4) = 0

This gives us two possible values for the common ratio r:
Either 4r + 3 = 0 \implies r = -3/4 Or 3r + 4 = 0 \implies r = -4/3

Step 6: Find the three terms for both cases.
Remember our terms are \frac{a}{r}, \ a, \ ar and a = -1.

Case 1: When r = -3/4
First term = \frac{-1}{-3/4} = 4/3

Second term = -1

Third term = -1 \times \left(\frac{-3}{4}\right) = 3/4

The terms are: 4/3, -1, 3/4

Case 2: When r = -4/3
First term = \frac{-1}{-4/3} = 3/4

Second term = -1

Third term = -1 \times \left(\frac{-4}{3}\right) = 4/3

The terms are: 3/4, -1, 4/3

Final Answer: The common ratio is \boxed{ -3/4 \text{ or } -4/3 } .

The terms are \boxed{ 4/3, -1, 3/4 } or \boxed{ 3/4, -1, 4/3 } .

Teacher’s Tip: If you ever write down your sequence and the second case is just the first case written backwards, congratulations! You have correctly solved a “product is given” G.P. problem. Because r and 1/r are both valid ratios, the sequence just flips direction!

Q14. Find the sum of ‘n’ terms of the series: 0.5+0.55+0.555+........ n terms.

Solution:
Step 1: Set up the sum S_n and extract common factors.
S_n = 0.5 + 0.55 + 0.555 + \dots \text{ to } n \text{ terms}

Convert decimals to fractions for easier manipulation:
S_n = \frac{5}{10} + \frac{55}{100} + \frac{555}{1000} + \dots

S_n = 5 \left[ \frac{1}{10} + \frac{11}{100} + \frac{111}{1000} + \dots \right]

Step 2: Introduce a 9 to create a pattern.
Multiply and divide the bracket by 9:
S_n = \frac{5}{9} \left[ \frac{9}{10} + \frac{99}{100} + \frac{999}{1000} + \dots \right]

Step 3: Rewrite terms as (1 – something).
S_n = \frac{5}{9} \left[ \left(1 - \frac{1}{10}\right) + \left(1 - \frac{1}{100}\right) + \left(1 - \frac{1}{1000}\right) + \dots \text{ to } n \text{ terms} \right]

Step 4: Separate the sequence into two parts.
Group the 1s together, and the fractions together:
S_n = \frac{5}{9} \left[ (1 + 1 + 1 \dots n \text{ times}) - \left( \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + \dots \text{ to } n \text{ terms} \right) \right]

S_n = \frac{5}{9} \left[ n - \left( \frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + \dots + \frac{1}{10^n} \right) \right]

Step 5: Apply the G.P. sum formula.
The second bracket is a G.P. with a = 1/10, r = 1/10.

Sum of G.P. = a \frac{1 - r^n}{1 - r} = \frac{1}{10} \frac{1 - (1/10)^n}{1 - 1/10} = \frac{1}{10} \frac{1 - 10^{-n}}{9/10} = \frac{1}{9}(1 - 10^{-n})

Step 6: Substitute back and finalize.
S_n = \frac{5}{9} \left[ n - \frac{1}{9}(1 - 10^{-n}) \right]

S_n = \frac{5}{81} \left[ 9n - 1 + 10^{-n} \right]

Final Answer: \boxed{ \frac{5}{81} [ 9n - 1 + 10^{-n} ] }

Teacher’s Tip: This exact methodology (taking the number common, then multiplying/dividing by 9) is standard for all repeating digit series like 7+77+777 or 0.3+0.33+0.333. Memorize this workflow!

Section D: 5 Marks Question

Q15. If \theta_{1}, \theta_{2}, \theta_{3} \dots \theta_{n} are in AP whose common difference is d, then show that \sec \theta_{1} \sec \theta_{2} + \sec \theta_{2} \sec \theta_{3} + \dots + \sec \theta_{n-1} \sec \theta_{n} = \frac{\tan \theta_{n} - \tan \theta_{1}}{\sin d}.

Solution:
Step 1: Rewrite the series in terms of sine and cosine.

Let the series be denoted as S.
S = \frac{1}{\cos \theta_1 \cos \theta_2} + \frac{1}{\cos \theta_2 \cos \theta_3} + \dots + \frac{1}{\cos \theta_{n-1} \cos \theta_n}

Step 2: Multiply and divide the entire series by \sin d.
S = \frac{1}{\sin d} \left[ \frac{\sin d}{\cos \theta_1 \cos \theta_2} + \frac{\sin d}{\cos \theta_2 \cos \theta_3} + \dots \right]

Step 3: Express d in terms of angles. Since the angles are in A.P. with common difference d, we know:
d = \theta_2 - \theta_1 = \theta_3 - \theta_2 = \dots = \theta_n - \theta_{n-1}

Substitute these forms of d into the numerators of their respective terms:
S = \frac{1}{\sin d} \left[ \frac{\sin(\theta_2 - \theta_1)}{\cos \theta_1 \cos \theta_2} + \frac{\sin(\theta_3 - \theta_2)}{\cos \theta_2 \cos \theta_3} + \dots + \frac{\sin(\theta_n - \theta_{n-1})}{\cos \theta_{n-1} \cos \theta_n} \right]

Step 4: Expand the sine of difference.
Using \sin(A-B) = \sin A \cos B - \cos A \sin B:
First term bracket = \frac{\sin \theta_2 \cos \theta_1 - \cos \theta_2 \sin \theta_1}{\cos \theta_1 \cos \theta_2}

Splitting the fraction:
= \frac{\sin \theta_2 \cos \theta_1}{\cos \theta_1 \cos \theta_2} - \frac{\cos \theta_2 \sin \theta_1}{\cos \theta_1 \cos \theta_2} = \tan \theta_2 - \tan \theta_1

Step 5: Apply this to all terms (Telescoping Series).
S = \frac{1}{\sin d} [ (\tan \theta_2 - \tan \theta_1) + (\tan \theta_3 - \tan \theta_2) + \dots + (\tan \theta_n - \tan \theta_{n-1}) ]

Step 6: Cancel intermediate terms.
Notice how +\tan \theta_2 cancels with -\tan \theta_2, +\tan \theta_3 cancels with -\tan \theta_3, etc.

Only the first negative term and the last positive term remain.
S = \frac{1}{\sin d} [ \tan \theta_n - \tan \theta_1 ]

\boxed{ \frac{\tan \theta_n - \tan \theta_1}{\sin d} } . Hence proved.

Teacher’s Tip: Whenever you see a series of reciprocals of trigonometric products, your first instinct should be a telescoping series. Multiplying and dividing by the sine of the angle difference is the skeleton key for these problems!

Section E: Case Study Questions (1 + 1 + 2)

Q16. In a library, 25 students are reading books on physics, chemistry, and mathematics. It was found that 15 students were reading mathematics, 12 reading physics and 11 reading chemistry, 5 students reading both mathematics and chemistry, 9 students reading both physics and mathematics, 4 students reading both physics and chemistry, and 3 students reading all three subjects.
(a) Find the number of students reading only Chemistry.
(b) Find the number of students reading only Mathematics.
(c) Find the number of students reading at least one of the subject and also find the number of students reading none of the subjects.

Solution:
Step 1: Extract data and define the variables.
Total Students, n(U) = 25.
Maths, n(M) = 15
Physics, n(P) = 12
Chemistry, n(C) = 11

Maths & Chem, n(M \cap C) = 5
Physics & Maths, n(P \cap M) = 9
Physics & Chem, n(P \cap C) = 4

All three, n(M \cap P \cap C) = 3

Step 2: Solve part (a) – Only Chemistry.
Formula for Only C = n(C) - n(P \cap C) - n(M \cap C) + n(M \cap P \cap C)

= 11 - 4 - 5 + 3 = 14 - 9 = 5

Step 3: Solve part (b) – Only Mathematics.
Formula for Only M = n(M) - n(P \cap M) - n(M \cap C) + n(M \cap P \cap C)

= 15 - 9 - 5 + 3 = 18 - 14 = 4

Step 4: Solve part (c) – At least one subject, and none.
“At least one” is the union of all three sets: n(M \cup P \cup C).

Using the formula:
n(M \cup P \cup C) = n(M) + n(P) + n(C) - n(M \cap P) - n(P \cap C) - n(M \cap C) + n(M \cap P \cap C)

= 15 + 12 + 11 - 9 - 4 - 5 + 3
= 38 - 18 + 3 = 23

Number of students reading NONE = Total Students – Students reading at least one

= n(U) - n(M \cup P \cup C) = 25 - 23 = 2

Final Answer:
(a) Only Chemistry = \boxed{ 5 }

(b) Only Mathematics = \boxed{ 4 }

(c) At least one = \boxed{ 23 } , None = \boxed{ 2 }

Teacher’s Tip: While formulas work well, drawing a Venn Diagram for these 3-set problems guarantees you won’t make an arithmetic mistake. Start by placing the ‘3’ in the center intersection and work your way outwards!

Q17. In Rangoli competition in school, Preeti made Rangoli in the equilateral shape. Each side of an equilateral triangle is 24 cm. The mid-point of its sides are joined to form another triangle. This process is going continuously infinite.
(a) Find the side of the 5th triangle is (in cm)
(b) Find the sum of perimeter of all triangle is (in cm)
(c) Find the area of all the triangle is (in sq cm).

Solution:
Step 1: Understand the geometric sequence formed by the sides.
When midpoints of an equilateral triangle are joined, the new triangle formed has sides exactly half the length of the original.

So, the sides form a G.P.:
1st triangle side, a_1 = 24
2nd triangle side, a_2 = 12
3rd triangle side, a_3 = 6

This is a G.P. with first term a = 24 and common ratio r = 1/2.

Step 2: Solve part (a) – Side of the 5th triangle.
Using the n-th term formula a_n = a r^{n-1}:

a_5 = 24 \left(\frac{1}{2}\right)^4 = 24 \times \frac{1}{16} = \frac{24}{16} = 1.5 cm.

Step 3: Solve part (b) – Sum of perimeters of all triangles.
Perimeter = 3 \times \text{side}.

Perimeters form a sequence: 3(24), 3(12), 3(6) \dots = 72, 36, 18 \dots

This is an infinite G.P. with a = 72 and r = 1/2.

Sum of infinite G.P. S_\infty = \frac{a}{1 - r}

S_\infty = \frac{72}{1 - 1/2} = \frac{72}{1/2} = 144 cm.

Step 4: Solve part (c) – Sum of areas of all triangles.
Area of an equilateral triangle = \frac{\sqrt{3}}{4} (\text{side})^2.

Area 1 = \frac{\sqrt{3}}{4} (24)^2 = 144\sqrt{3}

Area 2 = \frac{\sqrt{3}}{4} (12)^2 = 36\sqrt{3}

The areas form a G.P.: 144\sqrt{3}, 36\sqrt{3}, 9\sqrt{3} \dots

Here, a = 144\sqrt{3} and the common ratio r = \frac{36\sqrt{3}}{144\sqrt{3}} = \frac{1}{4}.

Sum of infinite G.P. S_\infty = \frac{a}{1 - r}

S_\infty = \frac{144\sqrt{3}}{1 - 1/4} = \frac{144\sqrt{3}}{3/4} = 144\sqrt{3} \times \frac{4}{3} = 48 \times 4\sqrt{3} = 192\sqrt{3} sq cm.

Final Answer:

(a) Side of 5th triangle = \boxed{ 1.5 \text{ cm} }

(b) Sum of perimeters = \boxed{ 144 \text{ cm} }

(c) Sum of areas = \boxed{ 192\sqrt{3} \text{ sq cm} }

Teacher’s Tip: Whenever dimensions (length) shrink by a ratio r, the perimeters also shrink by ratio r, but the areas will always shrink by ratio r^2! Notice how the area ratio was 1/4 while the side ratio was 1/2. Keep this golden geometric rule in mind!