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JEE Main Previous Year Questions (2026): Sequences and Series
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Page 1 JEE Main Previous Year Questions (2025): Sequences and Series Q1: The roots of the quadratic equation ?? ?? ?? - ???? + ?? = ?? are ???? th and ???? th terms of an arithmetic progression with common difference ?? ?? . If the sum of the first 11 terms of this arithmetic progression is ???? , then ?? - ?? ?? is equal to ____ . JEE Main 2025 (Online) 23rd January Evening Shift Ans: 474 Solution: ?? 11 = 11 2 ( 2?? + 10?? )= 88 ?? + 5?? = 8 ?? = 8 - 5 × 3 2 = 1 2 Roots are T 10 = a + 9 d = 1 2 + 9 × 3 2 = 14 T 11 = a + 10 d = 1 2 + 10 × 3 2 = 31 2 p 3 = T 10 + T 11 = 14 + 31 2 = 59 2 p = 177 2 q 3 = T 10 × T 11 = 7 × 31 = 217 q = 651 q - 2p = 651 - 177 = 474 Q2: The interior angles of a polygon with ?? sides, are in an A.P. with common difference ?? ° . If the largest interior angle of the polygon is ?????? ° , then ?? is equal to ____. Ans: 20 Solution: n 2 ( 2a + ( n - 1) 6)= ( n - 2)· 180 ° ( 1) an +3n 2 - 3n = ( n - 2)· 180 ° Page 2 JEE Main Previous Year Questions (2025): Sequences and Series Q1: The roots of the quadratic equation ?? ?? ?? - ???? + ?? = ?? are ???? th and ???? th terms of an arithmetic progression with common difference ?? ?? . If the sum of the first 11 terms of this arithmetic progression is ???? , then ?? - ?? ?? is equal to ____ . JEE Main 2025 (Online) 23rd January Evening Shift Ans: 474 Solution: ?? 11 = 11 2 ( 2?? + 10?? )= 88 ?? + 5?? = 8 ?? = 8 - 5 × 3 2 = 1 2 Roots are T 10 = a + 9 d = 1 2 + 9 × 3 2 = 14 T 11 = a + 10 d = 1 2 + 10 × 3 2 = 31 2 p 3 = T 10 + T 11 = 14 + 31 2 = 59 2 p = 177 2 q 3 = T 10 × T 11 = 7 × 31 = 217 q = 651 q - 2p = 651 - 177 = 474 Q2: The interior angles of a polygon with ?? sides, are in an A.P. with common difference ?? ° . If the largest interior angle of the polygon is ?????? ° , then ?? is equal to ____. Ans: 20 Solution: n 2 ( 2a + ( n - 1) 6)= ( n - 2)· 180 ° ( 1) an +3n 2 - 3n = ( n - 2)· 180 ° Now according to Q ?? + ( ?? - 1) 6 ° = 219 ° ? ?? = 225 ° - 6?? ° ( 2) Putting value of a from equation (2) in (1) We get ( 225?? - 6?? 2 )+ 3?? 2 - 3?? = 180?? - 360 ? 2n 2 - 42n- 360 = 0 ? n2- 14n- 120 = 0 n = 20, -6 (rejected) Q3: Let ?? ?? , ?? ?? , … , ?? ???????? be an Arithmetic Progression such that ?? ?? + ( ?? ?? + ?? ???? + ?? ???? + ? + ?? ???????? )+ ?? ???????? = ???????? . Then ?? ?? + ?? ?? + ?? ?? + ? + ?? ???????? is equal to ____ ? JEE Main 2025 (Online) 29th January Evening Shift Ans: 11132 Solution: a 1 + a 5 + a 10 + ? … + a 2020 + a 2024 = 2233 In an A.P. the sum of terms equidistant from ends is equal. ?? 1 + ?? 2024 = ?? 5 + ?? 2020 = ?? 10 + ?? 2015 ….. ? 203 pairs ? 203 ( ?? 1 + ?? 2024 )= 2233 Hence, S 2024 = 2024 2 ( a 1 + a 2024 ) = 1012× 11 = 11132 Q4: If the sum of the first ???? terms of the series ?? ·?? ?? +?? ·?? ?? + ?? ·?? ?? +?? ·?? ?? + ?? ·?? ?? +?? ·?? ?? + ?. is ?? ?? , where ?????? ( ?? , ?? )= ?? , then ?? + ?? is equal to ____ JEE Main 2025 (Online) 2nd April Evening Shift Ans: 441 Solution: 4.1 1+4.1 4 + 4.2 1+4.2 4 + 4.3 1+4.3 4 + ?. ?? ?? = 4?? 1 + 4?? 4 = 4?? 4?? 4 + 4?? 2 + 1 - 4?? 2 = 4?? ( 2?? 2 + 1) 2 - ( 2?? ) 2 ?? ?? = 4?? ( 2?? 2 - 2?? + 1) ( 2?? 2 + 2?? + 1) Page 3 JEE Main Previous Year Questions (2025): Sequences and Series Q1: The roots of the quadratic equation ?? ?? ?? - ???? + ?? = ?? are ???? th and ???? th terms of an arithmetic progression with common difference ?? ?? . If the sum of the first 11 terms of this arithmetic progression is ???? , then ?? - ?? ?? is equal to ____ . JEE Main 2025 (Online) 23rd January Evening Shift Ans: 474 Solution: ?? 11 = 11 2 ( 2?? + 10?? )= 88 ?? + 5?? = 8 ?? = 8 - 5 × 3 2 = 1 2 Roots are T 10 = a + 9 d = 1 2 + 9 × 3 2 = 14 T 11 = a + 10 d = 1 2 + 10 × 3 2 = 31 2 p 3 = T 10 + T 11 = 14 + 31 2 = 59 2 p = 177 2 q 3 = T 10 × T 11 = 7 × 31 = 217 q = 651 q - 2p = 651 - 177 = 474 Q2: The interior angles of a polygon with ?? sides, are in an A.P. with common difference ?? ° . If the largest interior angle of the polygon is ?????? ° , then ?? is equal to ____. Ans: 20 Solution: n 2 ( 2a + ( n - 1) 6)= ( n - 2)· 180 ° ( 1) an +3n 2 - 3n = ( n - 2)· 180 ° Now according to Q ?? + ( ?? - 1) 6 ° = 219 ° ? ?? = 225 ° - 6?? ° ( 2) Putting value of a from equation (2) in (1) We get ( 225?? - 6?? 2 )+ 3?? 2 - 3?? = 180?? - 360 ? 2n 2 - 42n- 360 = 0 ? n2- 14n- 120 = 0 n = 20, -6 (rejected) Q3: Let ?? ?? , ?? ?? , … , ?? ???????? be an Arithmetic Progression such that ?? ?? + ( ?? ?? + ?? ???? + ?? ???? + ? + ?? ???????? )+ ?? ???????? = ???????? . Then ?? ?? + ?? ?? + ?? ?? + ? + ?? ???????? is equal to ____ ? JEE Main 2025 (Online) 29th January Evening Shift Ans: 11132 Solution: a 1 + a 5 + a 10 + ? … + a 2020 + a 2024 = 2233 In an A.P. the sum of terms equidistant from ends is equal. ?? 1 + ?? 2024 = ?? 5 + ?? 2020 = ?? 10 + ?? 2015 ….. ? 203 pairs ? 203 ( ?? 1 + ?? 2024 )= 2233 Hence, S 2024 = 2024 2 ( a 1 + a 2024 ) = 1012× 11 = 11132 Q4: If the sum of the first ???? terms of the series ?? ·?? ?? +?? ·?? ?? + ?? ·?? ?? +?? ·?? ?? + ?? ·?? ?? +?? ·?? ?? + ?. is ?? ?? , where ?????? ( ?? , ?? )= ?? , then ?? + ?? is equal to ____ JEE Main 2025 (Online) 2nd April Evening Shift Ans: 441 Solution: 4.1 1+4.1 4 + 4.2 1+4.2 4 + 4.3 1+4.3 4 + ?. ?? ?? = 4?? 1 + 4?? 4 = 4?? 4?? 4 + 4?? 2 + 1 - 4?? 2 = 4?? ( 2?? 2 + 1) 2 - ( 2?? ) 2 ?? ?? = 4?? ( 2?? 2 - 2?? + 1) ( 2?? 2 + 2?? + 1) ?? ?? = ( 2?? 2 + 2?? + 1)- ( 2?? 2 - 2?? + 1) ( 2?? 2 - 2?? + 1) ( 2?? 2 + 2?? + 1) ?? ?? = ( 1 ?? 2 + ( ?? - 1) 2 - 1 ?? 2 + ( ?? + 1) 2 ) ? ?? =1 10 ??? ?? = ( 1 0 2 + 1 2 - 1 1 2 + 2 2 + 1 1 2 + 2 2 - 1 2 2 + 3 2 + ? 1 9 2 + 10 2 - 1 10 2 + 11 2 = 1 - 1 221 = 220 221 ? ?? + ?? = 220 + 221 = 441 Q5: Let ?? ?? , ?? ?? , ?? ?? , … be a G.P. of increasing positive terms. If ?? ?? ?? ?? = ???? and ?? ?? + ?? ?? = ???? , then ?? ?? is equal to: JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 812 B. 784 C. 628 D. 526 Ans: B Solution: First, let us denote the first term of the G.P. by ?? 1 = ?? and the common ratio (which is > 1, since the G.P. is increasing) by ?? . Then the terms are: ?? 1 = ?? , ?? 2 = ???? , ?? 3 = ?? ?? 2 , ?? 4 = ?? ?? 3 , ?? 5 = ?? ?? 4 , ?? 6 = ?? ?? 5 , … We are given: ?? 1 · ?? 5 = 28, i.e. ?? · ( ?? ?? 4 )= ?? 2 ?? 4 = 28. ?? 2 + ?? 4 = 29, i.e. ???? + ?? ?? 3 = ?? ( ?? + ?? 3 )= 29. We want to find ?? 6 = ?? ?? 5 . 1. Solve for ?? From (1): ?? 2 ?? 4 = 28 ? ?? 2 = 28 ?? 4 . From (2): ?? ( ?? + ?? 3 )= 29 ? ?? = 29 ?? +?? 3 . Page 4 JEE Main Previous Year Questions (2025): Sequences and Series Q1: The roots of the quadratic equation ?? ?? ?? - ???? + ?? = ?? are ???? th and ???? th terms of an arithmetic progression with common difference ?? ?? . If the sum of the first 11 terms of this arithmetic progression is ???? , then ?? - ?? ?? is equal to ____ . JEE Main 2025 (Online) 23rd January Evening Shift Ans: 474 Solution: ?? 11 = 11 2 ( 2?? + 10?? )= 88 ?? + 5?? = 8 ?? = 8 - 5 × 3 2 = 1 2 Roots are T 10 = a + 9 d = 1 2 + 9 × 3 2 = 14 T 11 = a + 10 d = 1 2 + 10 × 3 2 = 31 2 p 3 = T 10 + T 11 = 14 + 31 2 = 59 2 p = 177 2 q 3 = T 10 × T 11 = 7 × 31 = 217 q = 651 q - 2p = 651 - 177 = 474 Q2: The interior angles of a polygon with ?? sides, are in an A.P. with common difference ?? ° . If the largest interior angle of the polygon is ?????? ° , then ?? is equal to ____. Ans: 20 Solution: n 2 ( 2a + ( n - 1) 6)= ( n - 2)· 180 ° ( 1) an +3n 2 - 3n = ( n - 2)· 180 ° Now according to Q ?? + ( ?? - 1) 6 ° = 219 ° ? ?? = 225 ° - 6?? ° ( 2) Putting value of a from equation (2) in (1) We get ( 225?? - 6?? 2 )+ 3?? 2 - 3?? = 180?? - 360 ? 2n 2 - 42n- 360 = 0 ? n2- 14n- 120 = 0 n = 20, -6 (rejected) Q3: Let ?? ?? , ?? ?? , … , ?? ???????? be an Arithmetic Progression such that ?? ?? + ( ?? ?? + ?? ???? + ?? ???? + ? + ?? ???????? )+ ?? ???????? = ???????? . Then ?? ?? + ?? ?? + ?? ?? + ? + ?? ???????? is equal to ____ ? JEE Main 2025 (Online) 29th January Evening Shift Ans: 11132 Solution: a 1 + a 5 + a 10 + ? … + a 2020 + a 2024 = 2233 In an A.P. the sum of terms equidistant from ends is equal. ?? 1 + ?? 2024 = ?? 5 + ?? 2020 = ?? 10 + ?? 2015 ….. ? 203 pairs ? 203 ( ?? 1 + ?? 2024 )= 2233 Hence, S 2024 = 2024 2 ( a 1 + a 2024 ) = 1012× 11 = 11132 Q4: If the sum of the first ???? terms of the series ?? ·?? ?? +?? ·?? ?? + ?? ·?? ?? +?? ·?? ?? + ?? ·?? ?? +?? ·?? ?? + ?. is ?? ?? , where ?????? ( ?? , ?? )= ?? , then ?? + ?? is equal to ____ JEE Main 2025 (Online) 2nd April Evening Shift Ans: 441 Solution: 4.1 1+4.1 4 + 4.2 1+4.2 4 + 4.3 1+4.3 4 + ?. ?? ?? = 4?? 1 + 4?? 4 = 4?? 4?? 4 + 4?? 2 + 1 - 4?? 2 = 4?? ( 2?? 2 + 1) 2 - ( 2?? ) 2 ?? ?? = 4?? ( 2?? 2 - 2?? + 1) ( 2?? 2 + 2?? + 1) ?? ?? = ( 2?? 2 + 2?? + 1)- ( 2?? 2 - 2?? + 1) ( 2?? 2 - 2?? + 1) ( 2?? 2 + 2?? + 1) ?? ?? = ( 1 ?? 2 + ( ?? - 1) 2 - 1 ?? 2 + ( ?? + 1) 2 ) ? ?? =1 10 ??? ?? = ( 1 0 2 + 1 2 - 1 1 2 + 2 2 + 1 1 2 + 2 2 - 1 2 2 + 3 2 + ? 1 9 2 + 10 2 - 1 10 2 + 11 2 = 1 - 1 221 = 220 221 ? ?? + ?? = 220 + 221 = 441 Q5: Let ?? ?? , ?? ?? , ?? ?? , … be a G.P. of increasing positive terms. If ?? ?? ?? ?? = ???? and ?? ?? + ?? ?? = ???? , then ?? ?? is equal to: JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 812 B. 784 C. 628 D. 526 Ans: B Solution: First, let us denote the first term of the G.P. by ?? 1 = ?? and the common ratio (which is > 1, since the G.P. is increasing) by ?? . Then the terms are: ?? 1 = ?? , ?? 2 = ???? , ?? 3 = ?? ?? 2 , ?? 4 = ?? ?? 3 , ?? 5 = ?? ?? 4 , ?? 6 = ?? ?? 5 , … We are given: ?? 1 · ?? 5 = 28, i.e. ?? · ( ?? ?? 4 )= ?? 2 ?? 4 = 28. ?? 2 + ?? 4 = 29, i.e. ???? + ?? ?? 3 = ?? ( ?? + ?? 3 )= 29. We want to find ?? 6 = ?? ?? 5 . 1. Solve for ?? From (1): ?? 2 ?? 4 = 28 ? ?? 2 = 28 ?? 4 . From (2): ?? ( ?? + ?? 3 )= 29 ? ?? = 29 ?? +?? 3 . Plug ?? from (2) into (1). After some algebra (or by a systematic approach), one finds that ?? 2 = 28 ? ?? = v 28 = 2v 7 (since ?? > 1, we take the positive root). 2. Solve for ?? From (2), using ?? = 2v 7 : ?? + ?? 3 = 2v 7 + ( 2v 7) 3 = 2v 7 + 56v 7 = 58v 7. Hence, ?? = 29 58v 7 = 1 2v 7 . 3. Find ?? 6 ?? 6 = ?? ?? 5 = 1 2v 7 × ( 2v 7) 5 . Compute ( 2v 7) 5 . First, ( 2v 7) 2 = 28; thus ( 2v 7) 5 = ( 2v 7) 4 · ( 2v 7)= 784 × ( 2v 7)= 1568v 7. Therefore, ?? 6 = 1 2v 7 · 1568v 7 = 1568 2 ( since v 7 cancels )= 784. Ans: ?? 6 = 784. Hence, the correct option is Option B. Q6: Suppose that the number of terms in an A.P. is ?? ?? , ?? ? ?? . If the sum of all odd terms of the A.P. is 40 , the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27 , then k is equal to: JEE Main 2025 (Online) 22nd January Evening Shift Options: A. 8 B. 6 C. 4 D. 5 Ans: D Solution: ?? 1 , ?? 2 , ?? 3 , … … , ?? 2?? ? A.P. ? r=1 k ?a 2r-1 = 40, ? r=1 k ?a 2r = 55, a 2k - a 1 = 27 k 2 [2a 1 + ( k - 1) 2 d] = 40, k 2 [2a 2 + ( k - 1) 2 d] = 55, d = 27 2k - 1 a 1 = 40 k - ( k - 1) d = 55 k - kd d = 15 k ? 27 2k - 1 = 15 k ? 9k = 10k- 5 ? k = 5 Page 5 JEE Main Previous Year Questions (2025): Sequences and Series Q1: The roots of the quadratic equation ?? ?? ?? - ???? + ?? = ?? are ???? th and ???? th terms of an arithmetic progression with common difference ?? ?? . If the sum of the first 11 terms of this arithmetic progression is ???? , then ?? - ?? ?? is equal to ____ . JEE Main 2025 (Online) 23rd January Evening Shift Ans: 474 Solution: ?? 11 = 11 2 ( 2?? + 10?? )= 88 ?? + 5?? = 8 ?? = 8 - 5 × 3 2 = 1 2 Roots are T 10 = a + 9 d = 1 2 + 9 × 3 2 = 14 T 11 = a + 10 d = 1 2 + 10 × 3 2 = 31 2 p 3 = T 10 + T 11 = 14 + 31 2 = 59 2 p = 177 2 q 3 = T 10 × T 11 = 7 × 31 = 217 q = 651 q - 2p = 651 - 177 = 474 Q2: The interior angles of a polygon with ?? sides, are in an A.P. with common difference ?? ° . If the largest interior angle of the polygon is ?????? ° , then ?? is equal to ____. Ans: 20 Solution: n 2 ( 2a + ( n - 1) 6)= ( n - 2)· 180 ° ( 1) an +3n 2 - 3n = ( n - 2)· 180 ° Now according to Q ?? + ( ?? - 1) 6 ° = 219 ° ? ?? = 225 ° - 6?? ° ( 2) Putting value of a from equation (2) in (1) We get ( 225?? - 6?? 2 )+ 3?? 2 - 3?? = 180?? - 360 ? 2n 2 - 42n- 360 = 0 ? n2- 14n- 120 = 0 n = 20, -6 (rejected) Q3: Let ?? ?? , ?? ?? , … , ?? ???????? be an Arithmetic Progression such that ?? ?? + ( ?? ?? + ?? ???? + ?? ???? + ? + ?? ???????? )+ ?? ???????? = ???????? . Then ?? ?? + ?? ?? + ?? ?? + ? + ?? ???????? is equal to ____ ? JEE Main 2025 (Online) 29th January Evening Shift Ans: 11132 Solution: a 1 + a 5 + a 10 + ? … + a 2020 + a 2024 = 2233 In an A.P. the sum of terms equidistant from ends is equal. ?? 1 + ?? 2024 = ?? 5 + ?? 2020 = ?? 10 + ?? 2015 ….. ? 203 pairs ? 203 ( ?? 1 + ?? 2024 )= 2233 Hence, S 2024 = 2024 2 ( a 1 + a 2024 ) = 1012× 11 = 11132 Q4: If the sum of the first ???? terms of the series ?? ·?? ?? +?? ·?? ?? + ?? ·?? ?? +?? ·?? ?? + ?? ·?? ?? +?? ·?? ?? + ?. is ?? ?? , where ?????? ( ?? , ?? )= ?? , then ?? + ?? is equal to ____ JEE Main 2025 (Online) 2nd April Evening Shift Ans: 441 Solution: 4.1 1+4.1 4 + 4.2 1+4.2 4 + 4.3 1+4.3 4 + ?. ?? ?? = 4?? 1 + 4?? 4 = 4?? 4?? 4 + 4?? 2 + 1 - 4?? 2 = 4?? ( 2?? 2 + 1) 2 - ( 2?? ) 2 ?? ?? = 4?? ( 2?? 2 - 2?? + 1) ( 2?? 2 + 2?? + 1) ?? ?? = ( 2?? 2 + 2?? + 1)- ( 2?? 2 - 2?? + 1) ( 2?? 2 - 2?? + 1) ( 2?? 2 + 2?? + 1) ?? ?? = ( 1 ?? 2 + ( ?? - 1) 2 - 1 ?? 2 + ( ?? + 1) 2 ) ? ?? =1 10 ??? ?? = ( 1 0 2 + 1 2 - 1 1 2 + 2 2 + 1 1 2 + 2 2 - 1 2 2 + 3 2 + ? 1 9 2 + 10 2 - 1 10 2 + 11 2 = 1 - 1 221 = 220 221 ? ?? + ?? = 220 + 221 = 441 Q5: Let ?? ?? , ?? ?? , ?? ?? , … be a G.P. of increasing positive terms. If ?? ?? ?? ?? = ???? and ?? ?? + ?? ?? = ???? , then ?? ?? is equal to: JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 812 B. 784 C. 628 D. 526 Ans: B Solution: First, let us denote the first term of the G.P. by ?? 1 = ?? and the common ratio (which is > 1, since the G.P. is increasing) by ?? . Then the terms are: ?? 1 = ?? , ?? 2 = ???? , ?? 3 = ?? ?? 2 , ?? 4 = ?? ?? 3 , ?? 5 = ?? ?? 4 , ?? 6 = ?? ?? 5 , … We are given: ?? 1 · ?? 5 = 28, i.e. ?? · ( ?? ?? 4 )= ?? 2 ?? 4 = 28. ?? 2 + ?? 4 = 29, i.e. ???? + ?? ?? 3 = ?? ( ?? + ?? 3 )= 29. We want to find ?? 6 = ?? ?? 5 . 1. Solve for ?? From (1): ?? 2 ?? 4 = 28 ? ?? 2 = 28 ?? 4 . From (2): ?? ( ?? + ?? 3 )= 29 ? ?? = 29 ?? +?? 3 . Plug ?? from (2) into (1). After some algebra (or by a systematic approach), one finds that ?? 2 = 28 ? ?? = v 28 = 2v 7 (since ?? > 1, we take the positive root). 2. Solve for ?? From (2), using ?? = 2v 7 : ?? + ?? 3 = 2v 7 + ( 2v 7) 3 = 2v 7 + 56v 7 = 58v 7. Hence, ?? = 29 58v 7 = 1 2v 7 . 3. Find ?? 6 ?? 6 = ?? ?? 5 = 1 2v 7 × ( 2v 7) 5 . Compute ( 2v 7) 5 . First, ( 2v 7) 2 = 28; thus ( 2v 7) 5 = ( 2v 7) 4 · ( 2v 7)= 784 × ( 2v 7)= 1568v 7. Therefore, ?? 6 = 1 2v 7 · 1568v 7 = 1568 2 ( since v 7 cancels )= 784. Ans: ?? 6 = 784. Hence, the correct option is Option B. Q6: Suppose that the number of terms in an A.P. is ?? ?? , ?? ? ?? . If the sum of all odd terms of the A.P. is 40 , the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27 , then k is equal to: JEE Main 2025 (Online) 22nd January Evening Shift Options: A. 8 B. 6 C. 4 D. 5 Ans: D Solution: ?? 1 , ?? 2 , ?? 3 , … … , ?? 2?? ? A.P. ? r=1 k ?a 2r-1 = 40, ? r=1 k ?a 2r = 55, a 2k - a 1 = 27 k 2 [2a 1 + ( k - 1) 2 d] = 40, k 2 [2a 2 + ( k - 1) 2 d] = 55, d = 27 2k - 1 a 1 = 40 k - ( k - 1) d = 55 k - kd d = 15 k ? 27 2k - 1 = 15 k ? 9k = 10k- 5 ? k = 5 Q7: If the first term of an A.P. is 3 and the sum of its first four terms is equal to one fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to JEE Main 2025 (Online) 23rd January Morning Shift Options: A. -120 B. -1200 C. -1080 D. -1020 Ans: C Solution: The first term, ?? = 3 Common difference, ?? The formula for the sum of the first ?? terms of an A.P. is: ?? ?? = ?? 2 [2?? + ( ?? - 1) ?? ] Given: ?? 4 = 1 5 ( ?? 8 - ?? 4 ) This implies: 5?? 4 = ?? 8 - ?? 4 ? 6?? 4 = ?? 8 Substituting the sum formulas: 6 · 4 2 [2 × 3 + ( 4 - 1) ?? ] = 8 2 [2 × 3 + ( 8 - 1) ?? ] Simplifying: 6 × 2[6 + 3?? ] = 4[6 + 7?? ] 12( 6 + 3?? )= 4( 6 + 7?? ) 72 + 36?? = 24 + 28?? 36?? - 28?? = 24 - 72 8?? = -48 ?? = -6 Now, to find ?? 20 : ?? 20 = 20 2 [2 × 3 + ( 20 - 1) ( -6) ] ?? 20 = 10[6 + 19 × ( -6) ] ?? 20 = 10[6 - 114] ?? 20 = 10 × ( -108) ?? 20 = -1080 Thus, the sum of the first 20 terms is -1080 . Q8: Let ?? ?? = ?? ?? + ?? ?? + ?? ???? + ?? ???? + ? upto ?? terms. If the sum of the first six terms of an A.P. with first term -p and common difference ?? is v???????? ?? ???????? , then the absolute difference betwen ???? th and ???? th terms of the A.P. isRead More
FAQs on JEE Main Previous Year Questions (2026): Sequences and Series
| 1. What are sequences and series in mathematics? |  |
Ans. Sequences and series are fundamental concepts in mathematics. A sequence is an ordered list of numbers, which can be finite or infinite. A series, on the other hand, is the sum of the terms of a sequence. For example, if a sequence is given by 1, 2, 3, ..., the corresponding series would be 1 + 2 + 3 + ... and so on. Understanding these concepts is essential for solving various mathematical problems, particularly in algebra and calculus.
| 2. How do you determine the sum of a finite arithmetic series? | |
Ans. To find the sum of a finite arithmetic series, you can use the formula S = n/2 × (a + l), where S is the sum, n is the number of terms, a is the first term, and l is the last term. Alternatively, if you know the first term (a), the common difference (d), and the number of terms (n), you can also use the formula S = n/2 × (2a + (n - 1)d).
| 3. What is the difference between convergence and divergence in series? | |
Ans. Convergence and divergence refer to the behavior of a series as the number of terms increases. A series is said to converge if the sum of its terms approaches a finite number as more terms are added. Conversely, a series diverges if the sum grows without bound or does not approach a specific value. Understanding whether a series converges or diverges is crucial for studying infinite series in calculus.
| 4. What are geometric series, and how do you find their sum? | |
Ans. A geometric series is a series of terms where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio (r). The sum of the first n terms of a finite geometric series can be calculated using the formula S = a(1 - rⁿ) / (1 - r), where S is the sum, a is the first term, r is the common ratio, and n is the number of terms. For an infinite geometric series where the absolute value of r is less than 1, the sum can be found using S = a / (1 - r).
| 5. How can sequences and series be applied in real-life situations? | |
Ans. Sequences and series have numerous applications in real life. For example, they are used in finance to calculate interest over time, in physics to model wave patterns, and in computer science for algorithm analysis. Additionally, sequences are essential in programming for data structure manipulation and in statistics for analyzing trends within data sets. Understanding these concepts helps in making informed decisions based on mathematical reasoning in various fields.
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