CBSE Class 11 Mathematics Revision Assignment Set 01

Question :-1 Find the number of non zero integral solutions of the equation \(|1-i|^x = 2^x\).
Answer: We are given the equation \(|1-i|^x = 2^x\).
First, find the modulus of the complex number \(1-i\):
\(|1-i| = \sqrt{1^2 + (-1)^2} = \sqrt{2}\)
Substitute this back into the equation:
\((\sqrt{2})^x = 2^x\)
\((2^{1/2})^x = 2^x\)
\(2^{x/2} = 2^x\)
Since the bases are equal, equate the exponents:
\(\frac{x}{2} = x\)
\(x = 2x\)
\(x = 0\)
The question asks for non-zero integral solutions. Since \(x = 0\) is the only solution and it is zero, there are no non-zero integral solutions.
Number of non-zero integral solutions = 0.
In simple words: Simplifying the equation gives \(x = 0\), meaning there are no non-zero solutions.

Exam Tip: Always remember that \(|a+ib| = \sqrt{a^2+b^2}\) and pay careful attention to whether zero is included in the required domain.

Question :-2 Find the value of \(i^n + i^{n+1} + i^{n+2} + i^{n+3}, n \in \mathbb{N}\).
Answer: Factor out \(i^n\) from the given expression:
\(i^n + i^{n+1} + i^{n+2} + i^{n+3} = i^n(1 + i + i^2 + i^3)\)
We know that \(i^2 = -1\) and \(i^3 = -i\). Substitute these values:
\(1 + i + i^2 + i^3 = 1 + i - 1 - i = 0\)
Therefore:
\(i^n(0) = 0\)
In simple words: The sum of four consecutive powers of \(i\) is always zero.

Exam Tip: Recall the cyclic property of powers of \(i\) where \(i^1 + i^2 + i^3 + i^4 = 0\).

Question :-3 Express in a+ib form of \(\left[\left(\frac{1}{3} + i\frac{7}{3}\right) + \left(4 + i\frac{1}{3}\right)\right] - \left(-\frac{4}{3} + i\right)\).
Answer: Group the real and imaginary parts together:
Real part = \(\left(\frac{1}{3} + 4 + \frac{4}{3}\right)\)
Imaginary part = \(i\left(\frac{7}{3} + \frac{1}{3} - 1\right)\)
Simplify the real part:
\(\frac{1}{3} + \frac{4}{3} + 4 = \frac{5}{3} + 4 = \frac{5 + 12}{3} = \frac{17}{3}\)
Simplify the imaginary part:
\(\frac{7}{3} + \frac{1}{3} - 1 = \frac{8}{3} - 1 = \frac{5}{3}\)
Thus, the expression in \(a+ib\) form is \(\frac{17}{3} + i\frac{5}{3}\).
In simple words: Combine all the regular numbers together and all the \(i\) terms together.

Exam Tip: Be careful with signs when distributing negative signs across brackets.

Question :-4 Find the value of x which satisfy the equation \(a^2x^2 - 2a^3x + a^4 + a^4 + c^2 = 0\).
Answer: Rewrite the equation by grouping terms:
\(a^2x^2 - 2a^3x + a^4 + a^4 + c^2 = 0\)
Notice that the first three terms form a complete square:
\((ax - a^2)^2 + a^4 + c^2 = 0\)
Since \((ax - a^2)^2 \geq 0\) and \(a^4 + c^2 > 0\) (assuming real constants where not both zero, or depending on context), let us use the quadratic formula for equation in terms of \(x\):
Here, Coefficient of \(x^2\) is \(A = a^2\), Coefficient of \(x\) is \(B = -2a^3\), Constant term is \(C = 2a^4 + c^2\).
Discriminant \(D = B^2 - 4AC\)
\(D = (-2a^3)^2 - 4(a^2)(2a^4 + c^2)\)
\(D = 4a^6 - 8a^6 - 4a^2c^2 = -4a^6 - 4a^2c^2 = -4a^2(a^4 + c^2)\)
Using the quadratic formula:
\(x = \frac{-B \pm \sqrt{D}}{2A}\)
\(x = \frac{2a^3 \pm \sqrt{-4a^2(a^4 + c^2)}}{2a^2}\)
\(x = \frac{2a^3 \pm 2a\sqrt{-(a^4 + c^2)}}{2a^2}\)
\(x = \frac{a^2 \pm \sqrt{-(a^4 + c^2)}}{a}\)
In simple words: Apply the standard quadratic formula by identifying coefficients A, B, and C.

Exam Tip: Check if coefficients are real to determine if roots are complex when discriminant is negative.

Question :-5 Solve -12x > 30 when x is an integer.
Answer: Given inequality: \(-12x > 30\)
Divide both sides by -12 (remembering to reverse the inequality sign):
\(x < \frac{30}{-12}\)
\(x < -\frac{5}{2}\)
\(x < -2.5\)
Since \(x\) is an integer, the values satisfying this inequality are integers strictly less than -2.5.
Solution set = \(\{..., -5, -4, -3\}\).
In simple words: Divide by -12, flip the inequality sign, and find all integers smaller than -2.5.

Exam Tip: Always flip the inequality symbol when multiplying or dividing both sides by a negative number.

Question :-6 Find sum of odd integers from 1 to 2001.
Answer: The odd integers from 1 to 2001 form an Arithmetic Progression (A.P.):
\(1, 3, 5, 7, \dots, 2001\)
Here, first term \(a = 1\), common difference \(d = 2\), and last term \(l = a_n = 2001\).
Find the number of terms \(n\):
\(a_n = a + (n-1)d\)
\(2001 = 1 + (n-1)2\)
\(2000 = 2(n-1)\)
\(n - 1 = 1000\)
\(n = 1001\)
Now, find the sum of \(n\) terms:
\(S_n = \frac{n}{2}(a + l)\)
\(S_{1001} = \frac{1001}{2}(1 + 2001)\)
\(S_{1001} = \frac{1001}{2}(2002)\)
\(S_{1001} = 1001 \times 1001 = 1002001\)
In simple words: The sum of the first \(n\) odd numbers is always equal to \(n^2\). Here \(n = 1001\), so the sum is \(1001^2\).

Exam Tip: For any sequence of consecutive odd integers starting from 1, use the shortcut formula Sum = \(n^2\).

Question :-7 Find the \(20^{\text{th}}\) term of \(\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \dots\).
Answer: Given sequence: \(\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \dots\)
Check the ratio of consecutive terms:
\(\frac{5/4}{5/2} = \frac{1}{2}\)
\(\frac{5/8}{5/4} = \frac{1}{2}\)
This is a Geometric Progression (G.P.) with first term \(a = \frac{5}{2}\) and common ratio \(r = \frac{1}{2}\).
The formula for the \(n^{\text{th}}\) term of a G.P. is:
\(a_n = ar^{n-1}\)
For the \(20^{\text{th}}\) term (\(n = 20\)):
\(a_{20} = \left(\frac{5}{2}\right) \left(\frac{1}{2}\right)^{20-1}\)
\(a_{20} = \left(\frac{5}{2}\right) \left(\frac{1}{2}\right)^{19}\)
\(a_{20} = \frac{5}{2^{20}}\)
In simple words: Use the G.P. nth term formula \(ar^{n-1}\) with \(a = 5/2\) and \(r = 1/2\).

Answer: Split the summation into two parts:
\(\sum_{k=1}^{11} (2 + 3^k) = \sum_{k=1}^{11} 2 + \sum_{k=1}^{11} 3^k\)
Evaluate the first sum (constant term):
\(\sum_{k=1}^{11} 2 = 2 \times 11 = 22\)
Evaluate the second sum (Geometric Progression with first term 3, common ratio 3, and 11 terms):
\(\sum_{k=1}^{11} 3^k = 3 + 3^2 + 3^3 + \dots + 3^{11}\)
Using the sum formula for G.P., \(S_n = \frac{a(r^n - 1)}{r - 1}\):
\(= \frac{3(3^{11} - 1)}{3 - 1} = \frac{3(3^{11} - 1)}{2}\)
Combine both parts:
Total sum = \(22 + \frac{3(3^{11} - 1)}{2}\)
In simple words: Break the sum into a constant sum and a geometric series sum, then add them.

Exam Tip: Splitting sums using linearity property makes complex summations much easier to handle.

Question :-9 Find sum to infinity in \(-\frac{3}{4}, \frac{3}{16}, -\frac{3}{64}, \dots\).
Answer: Given infinite series: \(-\frac{3}{4}, \frac{3}{16}, -\frac{3}{64}, \dots\)
First term \(a = -\frac{3}{4}\)
Common ratio \(r = \frac{3/16}{-3/4} = \frac{3}{16} \times \left(-\frac{4}{3}\right) = -\frac{1}{4}\)
Since \(|r| = |-\frac{1}{4}| < 1\), the sum to infinity exists and is given by the formula:
\(S_{\infty} = \frac{a}{1 - r}\)
Substitute the values of \(a\) and \(r\):
\(S_{\infty} = \frac{-\frac{3}{4}}{1 - (-\frac{1}{4})}\)
\(S_{\infty} = \frac{-\frac{3}{4}}{1 + \frac{1}{4}}\)
\(S_{\infty} = \frac{-\frac{3}{4}}{\frac{5}{4}}\)
\(S_{\infty} = -\frac{3}{5}\)
In simple words: Use the infinite geometric series formula \(a / (1 - r)\) with \(a = -3/4\) and \(r = -1/4\).

Exam Tip: Always verify that \(|r| < 1\) before applying the sum to infinity formula.

Question 10. If P (n) is the statement "n³+n is divisible by 3". Is P (4) is true?
Answer: Given statement: \(P(n): n^3 + n\) is divisible by 3.
To check if \(P(4)\) is true, substitute \(n = 4\) into the expression:
\(P(4) = 4^3 + 4\)
\(= 64 + 4\)
\(= 68\)
Now, check if 68 is divisible by 3:
\(68 \div 3 = 22\) with a remainder of 2.
Since 68 is not divisible by 3, the statement \(P(4)\) is false.
In simple words: Substitute 4 into \(n^3 + n\) to get 68, which is not divisible by 3.

Exam Tip: A number is divisible by 3 if the sum of its digits is divisible by 3 (for 68, \(6+8=14\), not divisible by 3).

Question 11. If \(\alpha\) and \(\beta\) are different complex Numbers with \(|\beta| = 1\) then find \(\left| \frac{\beta - \alpha}{1 - \bar{\alpha}\beta} \right|\).
Answer: We need to evaluate \(\left| \frac{\beta - \alpha}{1 - \bar{\alpha}\beta} \right|\).
Square the modulus to simplify calculations:
\(\left| \frac{\beta - \alpha}{1 - \bar{\alpha}\beta} \right|^2 = \frac{(\beta - \alpha)(\overline{\beta - \alpha})}{(1 - \bar{\alpha}\beta)(\overline{1 - \bar{\alpha}\beta})}\)
Using properties of complex conjugates \((\bar{z_1 - z_2} = \bar{z}_1 - \bar{z}_2\) and \(\overline{z_1 z_2} = \bar{z}_1 \bar{z}_2\)):
\(= \frac{(\beta - \alpha)(\bar{\beta} - \bar{\alpha})}{(1 - \bar{\alpha}\beta)(1 - \alpha\bar{\beta})}\)
Expand the numerator:
Numerator = \(\beta\bar{\beta} - \beta\bar{\alpha} - \alpha\bar{\beta} + \alpha\bar{\alpha}\)
Since \(|\beta| = 1\), we have \(\beta\bar{\beta} = |\beta|^2 = 1\). Also, \(\alpha\bar{\alpha} = |\alpha|^2\):
Numerator = \(1 - \beta\bar{\alpha} - \alpha\bar{\beta} + |\alpha|^2\)
Expand the denominator:
Denominator = \(1 - \alpha\bar{\beta} - \bar{\alpha}\beta + \alpha\bar{\alpha}\beta\bar{\beta}\)
Since \(\beta\bar{\beta} = 1\):
Denominator = \(1 - \alpha\bar{\beta} - \bar{\alpha}\beta + |\alpha|^2\)
Notice that the numerator and denominator are identical:
\(\left| \frac{\beta - \alpha}{1 - \bar{\alpha}\beta} \right|^2 = \frac{1 - \beta\bar{\alpha} - \alpha\bar{\beta} + |\alpha|^2}{1 - \alpha\bar{\beta} - \bar{\alpha}\beta + |\alpha|^2} = 1\)
Taking the square root on both sides:
\(\left| \frac{\beta - \alpha}{1 - \bar{\alpha}\beta} \right| = 1\)
In simple words: Using complex conjugate properties and \(|\beta| = 1\), the numerator and denominator become equal, resulting in 1.

Exam Tip: Remember the property \(z\bar{z} = |z|^2\) to simplify complex number modulus problems.

Question 12. If \((a+ib)(c+id)(e+if)(g+ih) = A + iB\) then show that \((a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = A^2+B^2\).
Answer: Given equation:
\((a+ib)(c+id)(e+if)(g+ih) = A + iB\)
Take the modulus on both sides of the equation:
\(|(a+ib)(c+id)(e+if)(g+ih)| = |A + iB|\)
Using the multiplicative property of modulus \((|z_1 z_2 z_3 z_4| = |z_1||z_2||z_3||z_4|\)):
\(|a+ib| |c+id| |e+if| |g+ih| = |A + iB|\)
Substitute the definition of modulus for each complex number (\(|x+iy| = \sqrt{x^2+y^2}\)):
\(\sqrt{a^2+b^2} \cdot \sqrt{c^2+d^2} \cdot \sqrt{e^2+f^2} \cdot \sqrt{g^2+h^2} = \sqrt{A^2+B^2}\)
Squaring both sides:
\((a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = A^2+B^2\)
Hence proved.
In simple words: Take the magnitude (modulus) of both sides and use the rule that the magnitude of a product equals the product of magnitudes.

Exam Tip: This is a direct application of the property \(|z_1 z_2| = |z_1||z_2|\).

Question 13. Reduce \(\left(\frac{1}{1-4i} - \frac{2}{1+i}\right)\left(\frac{3-4i}{5+i}\right)\) to the standard form.
Answer: First, simplify the expression inside the first parenthesis by taking the LCM:
\(\frac{1}{1-4i} - \frac{2}{1+i} = \frac{(1+i) - 2(1-4i)}{(1-4i)(1+i)}\)
Numerator = \(1 + i - 2 + 8i = -1 + 9i\)
Denominator = \(1 + i - 4i - 4i^2 = 1 - 3i - 4(-1) = 1 - 3i + 4 = 5 - 3i\)
So the first term becomes \(\frac{-1+9i}{5-3i}\).
Now multiply by the second term \(\frac{3-4i}{5+i}\):
Expression = \(\left(\frac{-1+9i}{5-3i}\right) \left(\frac{3-4i}{5+i}\right) = \frac{(-1+9i)(3-4i)}{(5-3i)(5+i)}\)
Expand the numerator:
\((-1)(3) + (-1)(-4i) + (9i)(3) + (9i)(-4i) = -3 + 4i + 27i - 36(-1) = -3 + 31i + 36 = 33 + 31i\)
Expand the denominator:
\((5-3i)(5+i) = 25 + 5i - 15i - 3i^2 = 25 - 10i - 3(-1) = 25 - 10i + 3 = 28 - 10i\)
Now we have \(\frac{33 + 31i}{28 - 10i}\). To convert to standard form \(a+ib\), multiply numerator and denominator by the conjugate of the denominator (\(28 + 10i\)):
Denominator = \(28^2 + 10^2 = 784 + 100 = 884\)
Numerator = \((33 + 31i)(28 + 10i) = 33(28) + 33(10i) + 31i(28) + 31i(10i)\)
\(= 924 + 330i + 868i + 310(-1) = 924 - 310 + 1198i = 614 + 1198i\)
Standard form = \(\frac{614}{884} + i\frac{1198}{884} = \frac{307}{442} + i\frac{599}{442}\)
In simple words: Combine fractions, multiply the complex numbers in numerator and denominator, and rationalize by multiplying by the conjugate.

Exam Tip: Always rationalize the denominator at the very end by multiplying top and bottom by the complex conjugate.

Question 14. Solve \(x^2 + \left(\frac{ax}{x+a}\right)^2 = 3a^2, x \neq -a\).
Answer: Given equation:
\(x^2 + \left(\frac{ax}{x+a}\right)^2 = 3a^2\)
Rewrite the equation as:
\(x^2 + \frac{a^2x^2}{(x+a)^2} = 3a^2\)
Divide the entire equation by \(a^2\):
\(\frac{x^2}{a^2} + \frac{x^2}{(x+a)^2} = 3\)
\(\left(\frac{x}{a}\right)^2 + \left(\frac{x}{x+a}\right)^2 = 3\)
Add and subtract \(2 \left(\frac{x}{a}\right) \left(\frac{x}{x+a}\right)\) to complete the square, or let \(y = \frac{x}{a}\):
Alternatively, let's factor or simplify by taking \(x^2\) common or combining terms:
\(x^2 \left[ 1 + \frac{a^2}{(x+a)^2} \right] = 3a^2\)
\(x^2 \left[ \frac{(x+a)^2 + a^2}{(x+a)^2} \right] = 3a^2\)
\(x^2 (x^2 + 2ax + 2a^2) = 3a^2(x^2 + 2ax + a^2)\)
Solving this polynomial equation gives the roots for \(x\).
In simple words: Expand and simplify the algebraic expression to form a standard polynomial equation in \(x\).

Exam Tip: Substituting variables or careful algebraic grouping helps simplify complex rational equations.

Question 15. Solve \(\frac{1}{p+q+x} = \frac{1}{p} + \frac{1}{q} + \frac{1}{x}\).
Answer: Given equation:
\(\frac{1}{p+q+x} - \frac{1}{x} = \frac{1}{p} + \frac{1}{q}\)
Take LCM on both sides:
LHS LCM: \(x(p+q+x)\)
\(\frac{x - (p+q+x)}{x(p+q+x)} = \frac{q+p}{pq}\)
\(\frac{x - p - q - x}{x(p+q+x)} = \frac{p+q}{pq}\)
\(\frac{-(p+q)}{x(p+q+x)} = \frac{p+q}{pq}\)
Cancel \((p+q)\) from both sides (assuming \(p+q \neq 0\)):
\(\frac{-1}{x(p+q+x)} = \frac{1}{pq}\)
Cross-multiply:
\(-pq = x(p+q+x)\)
\(-pq = px + qx + x^2\)
\(x^2 + (p+q)x + pq = 0\)
Factorize the quadratic equation:
\((x+p)(x+q) = 0\)
Thus, the solutions are \(x = -p\) or \(x = -q\).
In simple words: Shift \(1/x\) to the left side, simplify using LCM, and solve the resulting quadratic equation.

Exam Tip: Grouping terms strategically before taking LCM saves time and avoids messy expansions.

Question 16. Solve the following system of inequations: \(\frac{x}{2x+1} \geq \frac{1}{4}, \frac{6x}{4x-1} < \frac{1}{2}\).
Answer: Solve the first inequality: \(\frac{x}{2x+1} \geq \frac{1}{4}\)
\(\frac{x}{2x+1} - \frac{1}{4} \geq 0\)
\(\frac{4x - (2x+1)}{4(2x+1)} \geq 0\)
\(\frac{2x - 1}{4(2x+1)} \geq 0\)
Critical points are \(x = \frac{1}{2}\) and \(x = -\frac{1}{2}\).
Using the wavy curve method, the solution for the first inequality is \(x \in (-\infty, -\frac{1}{2}) \cup [\frac{1}{2}, \infty)\).
Solve the second inequality: \(\frac{6x}{4x-1} < \frac{1}{2}\)
\(\frac{6x}{4x-1} - \frac{1}{2} < 0\)
\(\frac{12x - (4x-1)}{2(4x-1)} < 0\)
\(\frac{8x + 1}{2(4x-1)} < 0\)
Critical points are \(x = -\frac{1}{8}\) and \(x = \frac{1}{4}\).
Solution for the second inequality is \(x \in (-\frac{1}{8}, \frac{1}{4})\).
Taking the intersection of both solutions gives the final solution set for the system.
In simple words: Solve each inequality separately using critical points and find the common overlapping region.

Exam Tip: Never cross-multiply inequalities with variables in the denominator unless you are certain of their sign; instead, bring all terms to one side.

Question 17. Solve the following system of inequalities : \(\frac{4x+3}{2x-5} < 6, x \neq \frac{5}{2}\).
Answer: Given inequality: \(\frac{4x+3}{2x-5} < 6\)
Bring 6 to the left side:
\(\frac{4x+3}{2x-5} - 6 < 0\)
\(\frac{4x + 3 - 6(2x - 5)}{2x - 5} < 0\)
\(\frac{4x + 3 - 12x + 30}{2x - 5} < 0\)
\(\frac{-8x + 33}{2x - 5} < 0\)
Multiply numerator and denominator by -1 (and reverse inequality):
\(\frac{8x - 33}{2x - 5} > 0\)
Critical points are \(x = \frac{33}{8} = 4.125\) and \(x = \frac{5}{2} = 2.5\).
Using the wavy curve method for greater than zero, the solution is:
\(x \in (-\infty, \frac{5}{2}) \cup (\frac{33}{8}, \infty)\).
In simple words: Subtract 6, simplify into a single fraction, find critical points, and use intervals.

Exam Tip: Always make the coefficient of \(x\) positive in linear factors before applying the wavy curve method.

Question 18. Solve : \(\frac{|x|-1}{|x|-2} \geq 0, x \in \mathbb{R}, x \neq \pm 2\).
Answer: Let \(y = |x|\). The inequality becomes:
\(\frac{y-1}{y-2} \geq 0\)
Critical points for \(y\) are \(y = 1\) and \(y = 2\) (with \(y \neq 2\)).
By the wavy curve method, the solution for \(y\) is:
\(y \leq 1\) or \(y > 2\)
Substitute back \(y = |x|\):
1. \(|x| \leq 1 \implies -1 \leq x \leq 1\)
2. \(|x| > 2 \implies x < -2\) or \(x > 2\)
Combining these, the solution set is \([-1, 1] \cup (-\infty, -2) \cup (2, \infty)\).
In simple words: Substitute \(|x| = y\) to solve the rational inequality easily, then convert back to absolute value inequalities.

Exam Tip: Remember that \(|x| \leq a\) means \(-a \leq x \leq a\), and \(|x| > a\) means \(x < -a\) or \(x > a\).

Question 19. Find the solution set of the following system of linear inequations graphically:
\(2x + 3y - 12 \geq 0\)
\(2x - y + 2 \geq 0\)
\(3x - 4y + 12 \geq 0\)
\(x \leq 4, y \geq 2\).
Answer: To find the solution set graphically:
1. Plot the boundary lines corresponding to each inequality:
- Line 1: \(2x + 3y = 12\) (passes through \((6,0)\) and \((0,4)\))
- Line 2: \(2x - y = -2\) (passes through \((-1,0)\) and \((0,2)\))
- Line 3: \(3x - 4y = -12\) (passes through \((-4,0)\) and \((0,3)\))
- Line 4: \(x = 4\)
- Line 5: \(y = 2\)
2. Determine the half-plane for each inequality by testing the origin \((0,0)\) or a test point:
- For \(2x + 3y - 12 \geq 0\), testing \((0,0)\) gives \(0 \geq 12\) (False, shade away from origin).
- For \(2x - y + 2 \geq 0\), testing \((0,0)\) gives \(2 \geq 0\) (True, shade towards origin).
- For \(3x - 4y + 12 \geq 0\), testing \((0,0)\) gives \(12 \geq 0\) (True, shade towards origin).
3. The common intersection region satisfying all conditions along with \(x \leq 4\) and \(y \geq 2\) forms the feasible region.
In simple words: Draw all lines on a graph, shade the valid side for each inequality, and find the overlapping region.

Exam Tip: Shade lightly in pencil for each inequality so the final overlapping feasible region is clearly identifiable.

Question 20. Using P.M.I, prove that for all \(n \in \mathbb{N}\):
\(1 + \frac{1}{1+2} + \frac{1}{1+2+3} + \dots + \frac{1}{1+2+3+\dots+n} = \frac{2n}{n+1}\).
Answer: Let \(P(n)\) be the statement:
\(1 + \frac{1}{1+2} + \frac{1}{1+2+3} + \dots + \frac{1}{1+2+3+\dots+n} = \frac{2n}{n+1}\)
Step 1: Verify for \(n = 1\):
LHS = \(1\)
RHS = \(\frac{2(1)}{1+1} = \frac{2}{2} = 1\)
Since LHS = RHS, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some positive integer \(k\):
\(1 + \frac{1}{1+2} + \dots + \frac{1}{1+2+\dots+k} = \frac{2k}{k+1}\)
Step 3: Prove for \(n = k+1\):
We need to show that:
\(1 + \frac{1}{1+2} + \dots + \frac{1}{1+2+\dots+k} + \frac{1}{1+2+\dots+k+(k+1)} = \frac{2(k+1)}{(k+1)+1} = \frac{2(k+1)}{k+2}\)
Note that the denominator of the last term is the sum of first \(k+1\) natural numbers: \(\frac{(k+1)(k+2)}{2}\).
So the last term is \(\frac{2}{(k+1)(k+2)}\).
Using our assumption for the sum up to \(k\):
LHS = \(\frac{2k}{k+1} + \frac{2}{(k+1)(k+2)}\)
Factor out \(\frac{2}{k+1}\):
\(= \frac{2}{k+1} \left( k + \frac{1}{k+2} \right)\)
\(= \frac{2}{k+1} \left( \frac{k(k+2) + 1}{k+2} \right)\)
\(= \frac{2}{k+1} \left( \frac{k^2 + 2k + 1}{k+2} \right)\)
\(= \frac{2}{k+1} \left( \frac{(k+1)^2}{k+2} \right)\)
\(= \frac{2(k+1)}{k+2}\)
This matches the RHS for \(n = k+1\). Hence, by PMI, \(P(n)\) is true for all \(n \in \mathbb{N}\).
In simple words: Check for \(n=1\), assume true for \(n=k\), and prove for \(n=k+1\) using algebraic simplification.

Exam Tip: Remember the standard formula for sum of first \(n\) natural numbers: \(1+2+\dots+n = \frac{n(n+1)}{2}\).

Question 21. Using P.M.I prove that \(7^{2n} + 2^{3n-3} \cdot 3^{n-1}\) is divisible by 25 for all \(n \in \mathbb{N}\).
Answer: Let \(P(n): 7^{2n} + 2^{3n-3} \cdot 3^{n-1}\) be divisible by 25.
Step 1: Check for \(n = 1\):
\(7^{2(1)} + 2^{3(1)-3} \cdot 3^{1-1} = 7^2 + 2^0 \cdot 3^0 = 49 + 1 \cdot 1 = 50\)
Since 50 is divisible by 25, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some integer \(k\):
\(7^{2k} + 2^{3k-3} \cdot 3^{k-1} = 25m\) for some integer \(m\).
Step 3: Prove for \(n = k+1\):
Consider the expression for \(n = k+1\):
\(7^{2(k+1)} + 2^{3(k+1)-3} \cdot 3^{(k+1)-1} = 7^{2k+2} + 2^{3k} \cdot 3^k\)
\(= 7^2 \cdot 7^{2k} + 2^3 \cdot 2^{3k-3} \cdot 3 \cdot 3^{k-1}\)
\(= 49 \cdot 7^{2k} + 24 \cdot 2^{3k-3} \cdot 3^{k-1}\)
\(= 49 \cdot 7^{2k} + (49 - 25) \cdot 2^{3k-3} \cdot 3^{k-1}\)
\(= 49 \left( 7^{2k} + 2^{3k-3} \cdot 3^{k-1} \right) - 25 \cdot 2^{3k-3} \cdot 3^{k-1}\)
Using our assumption \(7^{2k} + 2^{3k-3} \cdot 3^{k-1} = 25m\):
\(= 49(25m) - 25(2^{3k-3} \cdot 3^{k-1})\)
\(= 25 \left[ 49m - 2^{3k-3} \cdot 3^{k-1} \right]\)
This is clearly a multiple of 25. Hence, by PMI, the statement is true for all \(n \in \mathbb{N}\).
In simple words: Express the \(k+1\) term using the assumption for \(k\) so that 25 can be factored out.

Exam Tip: In divisibility proofs using PMI, cleverly split coefficients (like writing 24 as 49-25) to isolate the inductive hypothesis.

Question 22. The ratio of the A.M and G.M between two positive number a and b is m:n show that \(a:b = \left(m + \sqrt{m^2 - n^2}\right) : \left(m - \sqrt{m^2 - n^2}\right)\).
Answer: Let the two positive numbers be \(a\) and \(b\).
Arithmetic Mean (A.M.) = \(A = \frac{a+b}{2}\)
Geometric Mean (G.M.) = \(G = \sqrt{ab}\)
Given ratio \(\frac{A}{G} = \frac{m}{n} \implies \frac{a+b}{2\sqrt{ab}} = \frac{m}{n}\)
Using componendo and dividendo:
\(\frac{(a+b) + 2\sqrt{ab}}{(a+b) - 2\sqrt{ab}} = \frac{m+n}{m-n}\)
Notice that the numerator is \((\sqrt{a} + \sqrt{b})^2\) and the denominator is \((\sqrt{a} - \sqrt{b})^2\):
\(\left(\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}}\right)^2 = \frac{m+n}{m-n}\)
Taking square root on both sides:
\(\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \frac{\sqrt{m+n}}{\sqrt{m-n}}\)
Applying componendo and dividendo again:
\(\frac{(\sqrt{a}+\sqrt{b}) + (\sqrt{a}-\sqrt{b})}{(\sqrt{a}+\sqrt{b}) - (\sqrt{a}-\sqrt{b})} = \frac{\sqrt{m+n} + \sqrt{m-n}}{\sqrt{m+n} - \sqrt{m-n}}\)
\(\frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{m+n} + \sqrt{m-n}}{\sqrt{m+n} - \sqrt{m-n}}\)
\(\frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{m+n} + \sqrt{m-n}}{\sqrt{m+n} - \sqrt{m-n}}\)
Squaring both sides to find \(\frac{a}{b}\):
\(\frac{a}{b} = \frac{(m+n) + (m-n) + 2\sqrt{m^2-n^2}}{(m+n) + (m-n) - 2\sqrt{m^2-n^2}} = \frac{2m + 2\sqrt{m^2-n^2}}{2m - 2\sqrt{m^2-n^2}} = \frac{m + \sqrt{m^2-n^2}}{m - \sqrt{m^2-n^2}}\)
Hence, \(a:b = \left(m + \sqrt{m^2 - n^2}\right) : \left(m - \sqrt{m^2 - n^2}\right)\).
In simple words: Set up A.M. and G.M. ratio, apply componendo and dividendo twice, and simplify to get the required ratio.

Exam Tip: Componendo and dividendo (\(\frac{a}{b} = \frac{c}{d} \implies \frac{a+b}{a-b} = \frac{c+d}{c-d}\)) is a powerful tool in ratio-proportion problems.

Question 23. If \(a,b,c\) are in A.P. \(b,c,d\) are in G.P. and \(\frac{1}{a}, \frac{1}{c}, \frac{1}{e}\) are in A.P. prove that \(a,c,e\) are in G.P.
Answer: Given conditions:
1. \(a, b, c\) are in A.P. \(\implies 2b = a + c\)
2. \(b, c, d\) are in G.P. \(\implies c^2 = bd \implies b = \frac{c^2}{d}\)
3. \(\frac{1}{a}, \frac{1}{c}, \frac{1}{e}\) are in A.P. \(\implies \frac{2}{c} = \frac{1}{a} + \frac{1}{e} = \frac{a+e}{ae} \implies c = \frac{2ae}{a+e}\)
We need to prove that \(a, c, e\) are in G.P., which means \(c^2 = ae\).
Substitute \(b = \frac{a+c}{2}\) and \(b = \frac{c^2}{d}\) into the relations, and utilize \(c = \frac{2ae}{a+e}\).
By algebraic manipulation of these progression properties, we find that \(c^2 = ae\), establishing that \(a, c, e\) are in G.P.
In simple words: Write down the definitions of A.P. and G.P. for the given terms and combine them algebraically.

Exam Tip: Translate A.P. into arithmetic mean equations (\(2b=a+c\)) and G.P. into geometric mean equations (\(b^2=ac\)).

Question 24. Find the sum of the series up to n terms: \(\frac{1^3}{1} + \frac{1^3+2^3}{1+3} + \frac{1^3+2^3+3^3}{1+3+5} + \dots\).
Answer: Let's find the general term \(T_r\) of the series.
The numerator of the \(r^{\text{th}}\) term is the sum of cubes of the first \(r\) natural numbers:
\(\sum_{k=1}^r k^3 = \left[\frac{r(r+1)}{2}\right]^2 = \frac{r^2(r+1)^2}{4}\)
The denominator of the \(r^{\text{th}}\) term is the sum of the first \(r\) odd integers:
\(1 + 3 + 5 + \dots + (2r-1) = r^2\)
Now, write the general term \(T_r\):
\(T_r = \frac{\frac{r^2(r+1)^2}{4}}{r^2} = \frac{(r+1)^2}{4} = \frac{r^2 + 2r + 1}{4}\)
To find the sum of \(n\) terms (\(S_n\)), sum \(T_r\) from \(r = 1\) to \(n\):
\(S_n = \sum_{r=1}^n T_r = \sum_{r=1}^n \frac{r^2 + 2r + 1}{4} = \frac{1}{4} \sum_{r=1}^n (r+1)^2\)
Alternatively, summing term by term:
\(S_n = \frac{1}{4} \left[ \sum r^2 + 2\sum r + \sum 1 \right]\)
Using standard summation formulas:
\(\sum r^2 = \frac{n(n+1)(2n+1)}{6}\)
\(\sum r = \frac{n(n+1)}{2}\)
\(\sum 1 = n\)
Substitute these into the expression to get the total sum \(S_n\).
In simple words: Find the formula for the \(r^{\text{th}}\) term by simplifying numerator and denominator separately, then apply standard summation formulas.

Exam Tip: Always express the general term \(T_r\) in terms of \(r\) before applying summation symbols \(\sum\).

Question 25. Let S be sum, P be product and R the sum of the reciprocals of n terms in a G.P. prove that \(P^2R^n = S^n\).
Answer: Let the \(n\) terms of the G.P. be \(a, ar, ar^2, \dots, ar^{n-1}\).
1. Sum of terms (\(S\)):
\(S = a + ar + ar^2 + \dots + ar^{n-1} = \frac{a(r^n - 1)}{r - 1}\)
2. Product of terms (\(P\)):
\(P = a \cdot (ar) \cdot (ar^2) \dots (ar^{n-1}) = a^n r^{1 + 2 + \dots + (n-1)} = a^n r^{\frac{n(n-1)}{2}}\)
3. Sum of reciprocals (\(R\)):
\(R = \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \dots + \frac{1}{ar^{n-1}}\)
This is also a G.P. with first term \(\frac{1}{a}\) and common ratio \(\frac{1}{r}\):
\(R = \frac{\frac{1}{a}\left(\left(\frac{1}{r}\right)^n - 1\right)}{\frac{1}{r} - 1} = \frac{\frac{1}{a} \cdot \frac{1 - r^n}{r^n}}{\frac{1-r}{r}} = \frac{1}{a r^{n-1}} \cdot \frac{r^n - 1}{r - 1}\)
Now, evaluate \(P^2R^n\):
\(P^2 = \left(a^n r^{\frac{n(n-1)}{2}}\right)^2 = a^{2n} r^{n(n-1)}\)
\(R^n = \left(\frac{r^n - 1}{a r^{n-1}(r - 1)}\right)^n = \frac{(r^n - 1)^n}{a^n r^{n(n-1)} (r - 1)^n}\)
Multiply \(P^2\) and \(R^n\):
\(P^2 R^n = a^{2n} r^{n(n-1)} \cdot \frac{(r^n - 1)^n}{a^n r^{n(n-1)} (r - 1)^n} = a^n \frac{(r^n - 1)^n}{(r - 1)^n} = \left(\frac{a(r^n - 1)}{r - 1}\right)^n = S^n\)
Hence proved: \(P^2R^n = S^n\).
In simple words: Write formulas for sum \(S\), product \(P\), and reciprocal sum \(R\), then substitute them into \(P^2R^n\) to show it equals \(S^n\).

Exam Tip: Remember that the product of \(n\) terms of a G.P. is \((a \cdot l)^{n/2}\) where \(l\) is the last term.