Top 50 Most Repeated SEQUENCES AND SERIES PYQs | JEE MAINS
A curated collection of the most important questions from SEQUENCES AND SERIES, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from SEQUENCES AND SERIES, fully solved with step-by-step concepts to prepare for JEE MAINS.
Standard formula: $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$....
Read Full Step-by-Step Solution →From $ar^4 = 81$, $ar = 24$, solve $r^3 = \frac{81}{24} = \frac{27}{8}$ → $r = \frac{3}{2}$, $a = 16$. Use $S_n = a\frac{r^n - 1}{r - 1}$...
Read Full Step-by-Step Solution →The AM–GM inequality states $\frac{a+b}{2} \ge \sqrt{ab}$, with equality only when $a=b$. Hence statements A and C are both correct, making option D t...
Read Full Step-by-Step Solution →$a_n = S_n - S_{n-1}$. Compute $S_{10} - S_9 = (300 + 50) - (243 + 45) = 350 - 288 = 62$? Wait: recalculate: $S_{10}=3(100)+50=350$, $S_9=3(81)+45=288...
Read Full Step-by-Step Solution →$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. Telescoping sum $= 1 - \frac{1}{101} = 0.99$...
Read Full Step-by-Step Solution →AM = $ \frac{a+b}{2} = 10 \Rightarrow a + b = 20 $. GM = $ \sqrt{ab} = 8 \Rightarrow ab = 64 $. Solving quadratic: $ x^2 - 20x + 64 = 0 \Rightarrow (x...
Read Full Step-by-Step Solution →Using $S_{n}=\frac{n}{2}[2a+(n-1)d]$, we get $S_{20}=10[2\times5+19\times3]=10[10+57]=10\times67=670$....
Read Full Step-by-Step Solution →HP terms correspond to AP reciprocals. Solving $\frac{1}{a+6d} = \frac{1}{13}, \frac{1}{a+12d} = \frac{1}{7}$ gives $a=1, d=2$....
Read Full Step-by-Step Solution →Use $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. Telescoping gives
Let $S = \sum_{n=1}^\infty (2n - 1) \left(\frac{1}{2}\right)^n$. Split: $S = 2\sum n r^n - \sum r^n$ with $r = 1/2$. Use $\sum_{n=1}^\infty r^n = \fra...
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Sequences and Series...
Read Full Step-by-Step Solution →Foundational check for Sequences and Series in Class 11. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Sequences and Series...
Read Full Step-by-Step Solution →Standard formula: $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$....
Read Full Step-by-Step Solution →From $ar^4 = 81$, $ar = 24$, solve $r^3 = \frac{81}{24} = \frac{27}{8}$ → $r = \frac{3}{2}$, $a = 16$. Use $S_n = a\frac{r^n - 1}{r - 1}$...
Read Full Step-by-Step Solution →The AM–GM inequality states $\frac{a+b}{2} \ge \sqrt{ab}$, with equality only when $a=b$. Hence statements A and C are both correct, making option D t...
Read Full Step-by-Step Solution →$a_n = S_n - S_{n-1}$. Compute $S_{10} - S_9 = (300 + 50) - (243 + 45) = 350 - 288 = 62$? Wait: recalculate: $S_{10}=3(100)+50=350$, $S_9=3(81)+45=288...
Read Full Step-by-Step Solution →$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. Telescoping sum $= 1 - \frac{1}{101} = 0.99$...
Read Full Step-by-Step Solution →AM = $ \frac{a+b}{2} = 10 \Rightarrow a + b = 20 $. GM = $ \sqrt{ab} = 8 \Rightarrow ab = 64 $. Solving quadratic: $ x^2 - 20x + 64 = 0 \Rightarrow (x...
Read Full Step-by-Step Solution →Using $S_{n}=\frac{n}{2}[2a+(n-1)d]$, we get $S_{20}=10[2\times5+19\times3]=10[10+57]=10\times67=670$....
Read Full Step-by-Step Solution →HP terms correspond to AP reciprocals. Solving $\frac{1}{a+6d} = \frac{1}{13}, \frac{1}{a+12d} = \frac{1}{7}$ gives $a=1, d=2$....
Read Full Step-by-Step Solution →Use $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. Telescoping gives
Let $S = \sum_{n=1}^\infty (2n - 1) \left(\frac{1}{2}\right)^n$. Split: $S = 2\sum n r^n - \sum r^n$ with $r = 1/2$. Use $\sum_{n=1}^\infty r^n = \fra...
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Sequences and Series...
Read Full Step-by-Step Solution →Foundational check for Sequences and Series in Class 11. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Sequences and Series...
Read Full Step-by-Step Solution →Standard formula: $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$....
Read Full Step-by-Step Solution →From $ar^4 = 81$, $ar = 24$, solve $r^3 = \frac{81}{24} = \frac{27}{8}$ → $r = \frac{3}{2}$, $a = 16$. Use $S_n = a\frac{r^n - 1}{r - 1}$...
Read Full Step-by-Step Solution →The AM–GM inequality states $\frac{a+b}{2} \ge \sqrt{ab}$, with equality only when $a=b$. Hence statements A and C are both correct, making option D t...
Read Full Step-by-Step Solution →$a_n = S_n - S_{n-1}$. Compute $S_{10} - S_9 = (300 + 50) - (243 + 45) = 350 - 288 = 62$? Wait: recalculate: $S_{10}=3(100)+50=350$, $S_9=3(81)+45=288...
Read Full Step-by-Step Solution →$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. Telescoping sum $= 1 - \frac{1}{101} = 0.99$...
Read Full Step-by-Step Solution →AM = $ \frac{a+b}{2} = 10 \Rightarrow a + b = 20 $. GM = $ \sqrt{ab} = 8 \Rightarrow ab = 64 $. Solving quadratic: $ x^2 - 20x + 64 = 0 \Rightarrow (x...
Read Full Step-by-Step Solution →Using $S_{n}=\frac{n}{2}[2a+(n-1)d]$, we get $S_{20}=10[2\times5+19\times3]=10[10+57]=10\times67=670$....
Read Full Step-by-Step Solution →HP terms correspond to AP reciprocals. Solving $\frac{1}{a+6d} = \frac{1}{13}, \frac{1}{a+12d} = \frac{1}{7}$ gives $a=1, d=2$....
Read Full Step-by-Step Solution →Use $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. Telescoping gives
Let $S = \sum_{n=1}^\infty (2n - 1) \left(\frac{1}{2}\right)^n$. Split: $S = 2\sum n r^n - \sum r^n$ with $r = 1/2$. Use $\sum_{n=1}^\infty r^n = \fra...
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Sequences and Series...
Read Full Step-by-Step Solution →Foundational check for Sequences and Series in Class 11. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Sequences and Series...
Read Full Step-by-Step Solution →Standard formula: $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$....
Read Full Step-by-Step Solution →From $ar^4 = 81$, $ar = 24$, solve $r^3 = \frac{81}{24} = \frac{27}{8}$ → $r = \frac{3}{2}$, $a = 16$. Use $S_n = a\frac{r^n - 1}{r - 1}$...
Read Full Step-by-Step Solution →The AM–GM inequality states $\frac{a+b}{2} \ge \sqrt{ab}$, with equality only when $a=b$. Hence statements A and C are both correct, making option D t...
Read Full Step-by-Step Solution →$a_n = S_n - S_{n-1}$. Compute $S_{10} - S_9 = (300 + 50) - (243 + 45) = 350 - 288 = 62$? Wait: recalculate: $S_{10}=3(100)+50=350$, $S_9=3(81)+45=288...
Read Full Step-by-Step Solution →$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. Telescoping sum $= 1 - \frac{1}{101} = 0.99$...
Read Full Step-by-Step Solution →AM = $ \frac{a+b}{2} = 10 \Rightarrow a + b = 20 $. GM = $ \sqrt{ab} = 8 \Rightarrow ab = 64 $. Solving quadratic: $ x^2 - 20x + 64 = 0 \Rightarrow (x...
Read Full Step-by-Step Solution →Using $S_{n}=\frac{n}{2}[2a+(n-1)d]$, we get $S_{20}=10[2\times5+19\times3]=10[10+57]=10\times67=670$....
Read Full Step-by-Step Solution →HP terms correspond to AP reciprocals. Solving $\frac{1}{a+6d} = \frac{1}{13}, \frac{1}{a+12d} = \frac{1}{7}$ gives $a=1, d=2$....
Read Full Step-by-Step Solution →