WebFirst of all, the arbitrary term should be 1/n·(n+4), not 1/n·(n+1). But okay, let's try to find the sum from n=1 to ∞ of 1/n·(n+4). We'll start by rewriting this with partial fractions. So we … 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 A.P , if Sn = n(4n + 1) , find the general form of an …
WebMay 5, 2024 · 1 answer If an AP is Sn = n (4n+1), then find the AP asked May 5, 2024 in Class X Maths by kabita (13.8k points) class-10 0 votes 1 answer Find the common difference of the AP 4,9,14,… If the first term changes to 6 and the common difference remains the same then write the new AP. asked Jan 20, 2024 in Class X Maths by priya … WebMar 29, 2024 · Given Sn = 4n – n2 Taking n = 1 S1 = 4 × 1−12 = 4 – 1 = 3 ∴ Sum of first term of AP is 3 Taking n = 2 in Sn S2 = 4×2−2^2 S2 = 8 – 4 S2 = 4 ∴ Sum of first 2 terms is 4 But … phone upto 25000
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Webs 1 = 4 (1) 2 − 1 = 4 − 1 = 3 Now, obviously, the sum of the first term will be the first term itself, as there are no other terms involved. When n = 2 , WebConsider an arithmetic progression (AP) whose first term is a 1 (or) a and the common difference is d.. The sum of first n terms of an arithmetic progression when the n th term is NOT known is S n = (n/2) [2a + (n - 1) d]; The sum of first n terms of an arithmetic progression when the n th term(a n) is known is S n = n/2[a 1 + a n]; Example: Mr. Kevin … WebGiven that sn = 4n^2 + 2n. ----- (1) Substitute n = 1 in (1), we get sn = 4(1)^2 + 2(1) = 4 + 2 = 6. So, Sum of the first term of AP is 6 i.e a = 6. Now, Substitute n = 2 in (1), we get sn = 4(2)^2 + 2(2) = 4 * 4 + 2 * 2 = 16 + 4 = 20. So, Sum of the first 2 terms = 20. Now, First-term + second term = 20 6 + a2 = 20 a2 = 20 - 6 a2 = 4. Hence in AP, phone upto 15000