WebSolution We know that, the nth term of an AP is; an= Sn−Sn−1 an= n(4n+1)−(n−1){4(n−1)+1} [∵ Sn= n(4n+1)] ⇒ an =4n2+n−(n−1)(4n−3) ⇒ an =4n2+n−4n2+3n+4n−3 ⇒ an =8n−3 P … WebFeb 1, 2024 · The sum of the first n terms of an AP is given by Sn, = 3n^2 - 4n. Determine the AP and the 12th term. ← Prev Question Next Question → +3 votes 79.8k views asked Feb 1, 2024 in Mathematics by Kundan kumar (51.5k points) The sum of the first n terms of an AP is given by S n, = 3n 2 - 4n. Determine the AP and the 12 th term. arithmetic progression

In an AP, if Sₙ = n(4n + 1), find the AP

WebMar 29, 2024 · Transcript. Ex 5.3, 3 In an AP (i) Given a = 5, d = 3, an = 50, find n and Sn. Given a = 5 , d = 3 , an = 50 We know that an = a + (n – 1) d Putting values 50 = 5 + (n – 1) ×3 50 = 5 + 3n – 3 50 = 2 + 3n 50 – 2 = 3n 48 = 3n 48/3=𝑛 n = 16 Now we need to find Sn Sn = 𝒏/𝟐 (𝟐𝒂+ (𝒏−𝟏)𝒅) Putting n = 16, a = 5, d = 3 ... WebSolution: The sum of n terms S n = 441 Similarly, S n-1 = 356 a = 13 d= n For an AP, S n = (n/2) [2a+ (n-1)d] Putting n = n-1 in above equation, l is the last term. It is also denoted by a n. The result obtained is: S n -S n-1 = a n So, 441-356 = a n a n = 85 = 13+ (n-1)d Since d=n, n (n-1) = 72 ⇒n 2 – n – 72= 0 Solving by factorization method, the place promised in our early days eng sub

If the sum of the first n terms of an AP is 4n - n - Cuemath

WebThe sum of the first n term of an A.P. is given by S n=3n 2+2n. Determine the A.P. and its 15 th term. Medium Solution Verified by Toppr S n=3n 2+2n Taking n=1, we get S 1=3(1) 2+2(1) ⇒S 1=3+2 ⇒S 1=5 ⇒a 1=5 Taking n=2, we get S 2=3(2) 2+2(2) ⇒S 2=12+4 ⇒S 2=16 ∴a 2=S 2−S 1=16−5=11 Taking n=3, we get S 3=3(3) 2+2(3) ⇒S 3=27+6 ⇒S 3=33 WebMar 31, 2024 · S n = n (4n + 1) Formula: a = first term d = common difference Calculation: S 1 = 1 (4 × 1 + 1) ⇒ S 1 = 4 + 1 = 5 S 2 = 2 (4 × 2 + 1) ⇒ S 2 = 2 × 9 = 18 Second term = S 2 – … the place recensione