CBSE 10th Grade Mathematics > Bonus QuestionsBonus QuestionsPrevious Question 1 of 31 / 3NextReveal all steps Given Sn = 2n² + 3n. Using an = Sn - S(n-1), First term: a1 = S1 = 2(1)² + 3(1) = 5. For n ≥ 2: an = Sn - S(n-1) = (2n² + 3n) - [2(n-1)² + 3(n-1)]. = 2n² + 3n - (2n² - 4n + 2 + 3n - 3). = 4n + 1. So nth term: an = 4n + 1. Check: a2 = 9, a3 = 13. Common difference d = 9 - 5 = 4. Or, using AP sum formula: Sn = n/2 [2a + (n-1)d]. Given Sn = 2n² + 3n. Equating: n/2 [2a + (n-1)d] = 2n² + 3n. n[2a + (n-1)d] = 4n² + 6n. 2a + (n-1)d = 4n + 6. Expanding: dn - d + 2a = 4n + 6. Compare coefficients: d = 4, -d + 2a = 6 ⇒ -4 + 2a = 6 ⇒ a = 5. So nth term: an = a + (n-1)d = 5 + (n-1)4 = 4n + 1. Direct substitution check: S1 = 5, S2 = 14, S3 = 27, S4 = 44. a1 = 5, a2 = 9, a3 = 13, a4 = 17. So sequence = 5, 9, 13, 17… which is an AP. Final Answer: First term = 5, Common difference = 4, nth term = 4n + 1.
