Sequences and Series Miscellaneous Exercise Class 11 NCERT

Deconstructing the Miscellaneous Exercise Challenges

The miscellaneous exercise is designed to develop mathematical lateral thinking. Instead of straightforward applications of standard formulas, you will encounter problems where sequences are nested, or where you must convert a given non-standard sequence into an AP or GP. A primary theme here is the relationship between different means—specifically, the AM-GM inequality, which states that for any positive real numbers, their Arithmetic Mean is always greater than or equal to their Geometric Mean ($AM \ge GM$). Another major component is the summation of special series using sigma ($\sum$) notation. You must be comfortable manipulating $\sum n = \frac{n(n+1)}{2}$, $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$, and $\sum n^3 = \left[\frac{n(n+1)}{2}\right]^2$. Success lies in representing the $r$-th term ($T_r$) of the series algebraically and then applying the summation operator term-by-term.

Step-by-Step Approach to Summing Special Series

1. Identify the General Term (Tr) — Examine the terms of the given series to find a pattern. Express the $r$-th term, $T_r$, as an algebraic function of $r$. For example, if the series is $1\cdot2 + 2\cdot3 + 3\cdot4 + \dots$, then $T_r = r(r+1) = r^2 + r$.
2. Apply the Sigma Operator — Express the sum of $n$ terms as $S_n = \sum_{r=1}^{n} T_r$. Distribute the sigma operator across the individual algebraic terms: $S_n = \sum r^2 + \sum r$.
3. Substitute Standard Summation Formulas — Replace the sigma expressions with their standard closed-form algebraic formulas. For instance, substitute $\sum r^2$ with $\frac{n(n+1)(2n+1)}{6}$ and $\sum r$ with $\frac{n(n+1)}{2}$.
4. Factorize and Simplify — Avoid expanding everything immediately. Instead, look for common algebraic factors (like $n(n+1)$) to simplify the expression efficiently and prevent complex arithmetic errors.

Critical Board Exam Traps & Tips

• The AM-GM Trap: Remember that the inequality $AM \ge GM$ (i.e., $\frac{a+b}{2} \ge \sqrt{ab}$) holds true only for positive real numbers. Do not apply this rule if the terms are not explicitly stated as positive.
• Algebraic Factoring: In special series summation, do not expand the cubic or quadratic expressions fully at the start. Always factor out common terms like $n$, $n+1$, or fractions first. This reduces calculation errors by 80%.
• Infinite GP Condition: When working with the sum of an infinite GP ($S_\infty = \frac{a}{1-r}$), ensure that $|r| < 1$. If $r \ge 1$, the sum to infinity does not exist.

Practice Questions with Solutions

• Q: Find the sum of $n$ terms of the sequence $7 + 77 + 777 + 7777 + \dots$ A: Step 1: Let $S_n = 7 + 77 + 777 + \dots$ to $n$ terms. Factor out 7: $S_n = 7(1 + 11 + 111 + \dots$ to $n$ terms) Step 2: Multiply and divide by 9 to create a base of 10: $S_n = \frac{7}{9}(9 + 99 + 999 + \dots$ to $n$ terms) Step 3: Rewrite each term as a power of 10: $S_n = \frac{7}{9}[(10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1)]$ Step 4: Group the GP terms and the constant terms separately: $S_n = \frac{7}{9}[(10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots$ to $n$ terms)] Step 5: Apply the GP sum formula $S_n = \frac{a(r^n - 1)}{r - 1}$ where $a = 10, r = 10$: $S_n = \frac{7}{9}[\frac{10(10^n - 1)}{10 - 1} - n]$ $S_n = \frac{7}{9}[\frac{10(10^n - 1)}{9} - n]$ Final answer: $S_n = \frac{7}{81}[10(10^n - 1) - 9n]$
• Q: If $a, b, c$ are in AP, $b, c, d$ are in GP, and $\frac{1}{c}, \frac{1}{d}, \frac{1}{e}$ are in AP, prove that $a, c, e$ are in GP. A: Step 1: Write down the mathematical implications of the given conditions: 1. Since $a, b, c$ are in AP $\implies 2b = a + c \implies b = \frac{a+c}{2}$ 2. Since $b, c, d$ are in GP $\implies c^2 = bd$ 3. Since $\frac{1}{c}, \frac{1}{d}, \frac{1}{e}$ are in AP $\implies \frac{2}{d} = \frac{1}{c} + \frac{1}{e} = \frac{c+e}{ce} \implies d = \frac{2ce}{c+e}$ Step 2: Substitute the values of $b$ and $d$ into the GP condition $c^2 = bd$: $c^2 = (\frac{a+c}{2}) (\frac{2ce}{c+e})$ Step 3: Simplify the equation by canceling 2 on the right-hand side: $c^2 = \frac{(a+c)ce}{c+e}$ Step 4: Divide both sides by $c$ (assuming $c \neq 0$): $c = \frac{(a+c)e}{c+e}$ Step 5: Cross-multiply and expand: $c(c+e) = e(a+c) \implies c^2 + ce = ae + ce$ Step 6: Subtract $ce$ from both sides: $c^2 = ae$ This proves that $a, c, e$ form a Geometric Progression. Final answer: Since $c^2 = ae$, $a, c, e$ are in GP.
• Q: Find the sum of the first $n$ terms of the series: $5 + 11 + 19 + 29 + 41 + \dots$ A: Step 1: Note that the differences between successive terms are $6, 8, 10, 12, \dots$, which form an AP. Let's use the difference method. $S_n = 5 + 11 + 19 + 29 + \dots + a_{n-1} + a_n$ Rewrite the series shifted by one position: $S_n = 0 + 5 + 11 + 19 + \dots + a_{n-1} + a_n$ Step 2: Subtract the second equation from the first: $0 = 5 + (11-5) + (19-11) + (29-19) + \dots + (a_n - a_{n-1}) - a_n$ $a_n = 5 + [6 + 8 + 10 + \dots \text{ to } (n-1) \text{ terms}]$ Step 3: Use the AP sum formula for the term in brackets ($a = 6, d = 2$ for $n-1$ terms): $a_n = 5 + \frac{n-1}{2}[2(6) + (n-2)2]$ $a_n = 5 + \frac{n-1}{2}[12 + 2n - 4] = 5 + \frac{n-1}{2}[2n + 8]$ $a_n = 5 + (n-1)(n+4) = 5 + n^2 + 3n - 4 = n^2 + 3n + 1$ Step 4: Now, find the sum $S_n = \sum_{r=1}^{n} a_r = \sum r^2 + 3\sum r + \sum 1$: $S_n = \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{2} + n$ Step 5: Factor out $\frac{n}{6}$ to simplify: $S_n = \frac{n}{6}[(n+1)(2n+1) + 9(n+1) + 6]$ $S_n = \frac{n}{6}[2n^2 + 3n + 1 + 9n + 9 + 6] = \frac{n}{6}[2n^2 + 12n + 16]$ $S_n = \frac{n(2)(n^2 + 6n + 8)}{6} = \frac{n(n+2)(n+4)}{3}$ Final answer: $S_n = \frac{n(n+2)(n+4)}{3}$
• Q: For any positive real numbers $a, b, c$, show that $(a+b)(b+c)(c+a) \ge 8abc$. A: Step 1: Recall the AM-GM inequality for two positive numbers $x$ and $y$: $\frac{x+y}{2} \ge \sqrt{xy} \implies (x+y) \ge 2\sqrt{xy}$ Step 2: Apply this inequality to the pairs $(a, b)$, $(b, c)$, and $(c, a)$ individually: 1. For positive real numbers $a$ and $b$: $a+b \ge 2\sqrt{ab}$ (Equation 1) 2. For positive real numbers $b$ and $c$: $b+c \ge 2\sqrt{bc}$ (Equation 2) 3. For positive real numbers $c$ and $a$: $c+a \ge 2\sqrt{ca}$ (Equation 3) Step 3: Since $a, b, c$ are positive, all terms are positive. Multiply Equation 1, Equation 2, and Equation 3 together: $(a+b)(b+c)(c+a) \ge (2\sqrt{ab})(2\sqrt{bc})(2\sqrt{ca})$ Step 4: Simplify the right side of the expression: Right Side = $8 \cdot \sqrt{a^2 b^2 c^2} = 8abc$ Step 5: Combine both sides to obtain the final inequality: $(a+b)(b+c)(c+a) \ge 8abc$ Final answer: Proved that $(a+b)(b+c)(c+a) \ge 8abc$ using the AM-GM inequality.

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