In an a.p if sn n 4n + 1 find the a.p
WebIn an AP, if s n =n (4n + 1), then find the AP. Solution: We know that, the n th term of an AP is Hence, the required AP is 5,13, 21,… Question 25: In an AP, if s n = 3n 2 + 5n and a k = 164, then find the value of k. Solution: Question 26: If s n denotes the sum of first n terms of an AP, then prove that s 12 =3(s 8-s 4) Solution: Question 27:
In an a.p if sn n 4n + 1 find the a.p
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WebSolution: Given, the expression for the sum of the terms is Sₙ = n (4n + 1) We have to find the AP. Put n = 1, S₁ = 1 (4 (1) + 1) = 4 + 1 = 5 Put n =2, S₂ = 2 (4 (2) + 1) = 2 (8 + 1) = 2 (9) = 18 … 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,
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 WebSince this holds for all n, we can plug in n=-1 and n=1 to get two equations: 1/-3=A/-1 +B/3 1/5=A/1+B/5 Clean up the fractions to get -1=-3A+B and 1=5A+B. Subtract the second …
WebAug 31, 2015 · The first five terms are #color(blue)(9,13,17,21,25# Explanation: #a_n=4n+5# We can find terms #1 # to #5# by substituting #n# respectively in the expression.. #a_n ... 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 …
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 …
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, fc 3095WebJan 20, 2014 · In an A.P.,if Sn = n (4n+1) ,find the A.P. Share with your friends 2 Follow 4 Varun.Rawat, Meritnation Expert added an answer, on 21/1/14 We have, S n = 4n 2 + n put n = 1, we get S 1 = 4 (1) 2 + 1 = 4 + 1 = 5 So, First term, a 1 = 5 Put n = 2 , we get S 2 = 4 (2) 2 + 2 = 18 so, a 2 = S 2 - S 1 = 18 - 5 = 13 Put n = 3, we get S 3 = 4 (3) 2 + 3 = 39 fc303机芯WebNCERT Exemplar Class 10 Maths Exercise 5.3 Sample Problem 1. If the numbers n - 2, 4n - 1 and 5n + 2 are in AP, find the value of n. Summary: An arithmetic progression is a sequence where each term, except the first term, is obtained by adding a fixed number to its previous term. If the numbers n - 2, 4n - 1 and 5n + 2 are in AP, then the value ... fc 3079308WebIn an AP, if S n=n(4n+1), fill the AP is 5, 13, __, --- Medium Solution Verified by Toppr Correct option is A) S n=n(4n+1) ∴S 1=a=1(4+1)=5 and S 2=a 1+a 2=2(4×2+1)=18 … fringe season 2 episode 3WebFeb 5, 2024 · answered If the sum of n terms of an AP is given by Sn=n (4n+1),then find the nth term of the AP See answers Advertisement ideba2011 Answer:given below Step-by-step explanation: follow the steps..... Advertisement sunitasahuo4 Answer: I think it will be help you Advertisement Advertisement fringe season 2 episode 2WebIn 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. fc3099461WebSep 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 fringe season 2 imdb