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, WebJan 28, 2024 · In an AP, if Sn = n (4n + 1), find the AP - YouTube #class10#arithmeticprogressionsIn an AP, if Sn = n (4n + 1), find the AP …
Arithmetic Progression - Formula, Definition, Examples, nth term
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 … 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 , short bed truck for sale near me
Partial sums: formula for nth term from partial sum
WebIn the given AP, the first term is a = 7 and the common difference is d = 4. Let us assume that 301 is the n th term of AP. Then: T n = a + (n - 1)d 301 = 7 + (n - 1) 4 301 = 7 + 4n - 4 301 = 4n + 3 298 = 4n n = 74.5 But 'n' must be an integer. Hence 301 cannot be a term of the given AP. Answer: 301 cannot be a term of the given AP. WebFeb 26, 2024 · how this formula is substituted in to the sum plzz explain step wise up to 8n-3 what u r thiking Advertisement Expert-Verified Answer 217 people found it helpful abhi178 … WebSep 20, 2024 · Sn =n ( 4n + 1 ) = 4n^2 + n we know that, Tn = Sn - S (n-1) =4n^2+n -4 (n-1)^2 - (n-1) =4 (n^2-n^2+2n-1)+ (n-n+1) =8n - 4 + 1 = 8n -3 hence , Tn = 8n -3 T1 =8 (1)-3 =5 T2= 8 (2)-3 =13 so, AP is 5, 13 , 21 and so on Find Math textbook solutions? Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 NCERT Class 9 Mathematics 619 solutions short bed truck for sale craigslist