In an a.p if sn n 4n + 1 find the a.p
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 … WebGiven sum of first n terms of the AP is Sn = 4n - n² Put n = 1, we get S1 = 4*1 - 1² = 4 – 1 = 3 So first term = 3 Now, sum of first two terms S2 = 4*2−2² (Put n=2) = 8−4 = 4 So sum of first two terms = 4 Therefore Second term =S2 −S1 =4−3 =1 So second term = 1 Again S3 = 4×3 - 3² (Put n= 3) = 12 – 9 = 3 Therefore Third term = S3 − S2 = 3 – 4 = – 1
In an a.p if sn n 4n + 1 find the a.p
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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. WebClass 10 Maths NCERT Solutions Chapter 5 Exercise 5.3 Question 11 . Summary: If the sum of the first n terms of an AP is 4n - n 2, then the first term is equal to 3, the sum of first two terms is equal to 4, the second term is equal to 1, and 3rd term, 10th term and the nth terms are equal to -1, -15, (5 - 2n) respectively.
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, WebLet 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 …
WebIf Sn, the sum of first n terms of an AP is given by Sn = 3n2-4n, Find the nth term? Nihar Hirani 20.9K subscribers Subscribe 29K views 2 years ago CBSE class X Arithmetic … WebSep 20, 2024 · Expert-Verified Answer 26 people found it helpful Wafabhatt given , 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
WebJan 11, 2016 · IN an AP, if sn=n [4n+1] find the AP Asked by archita123 11 Jan, 2016, 06:12: PM Expert Answer Answered by 11 Jan, 2016, 06:22: PM Application Videos This …
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