WebMar 16, 2024 · Answer: d = -2 Step-by-step explanation: Because the sum of first n terms of any AP is in the form (d/2)n² + (a - d/2)n, where d is the common difference and a is the first term. So that the coefficient of n² is half of the common difference. Here in S_n = 4n - n², the coefficient of n² is -1, so that half of common difference is -1. WebMar 30, 2024 · The sum of first n terms of an AP is given by S n = 2n 2 + 3n . Find the sixteenth term of the AP. Find the sixteenth term of the AP. This is a question of CBSE Sample Paper - Class 10 - 2024/18.
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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 ... WebLet n = 1, then a (1) = S (1) - S (0) and S (n) = (n+1)/ (n+10) that implies S (1) = 2/11,S (0) = 1/10 but S (0) = Sum of first 0 terms which is equal to zero ( S (0) = 0 ) and that is a contradiction. So the formula a (n) = S (n) -S (n-1) works only for n > 1.
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, WebIn an AP, if d = – 4, n = 7 and a n = 4, then a is equal to (a) 6 (b) 7 (c) 20 (d) 28 Solution: Question 2: In an AP, if a = 3.5, d = 0 and n = 101, then a n will be (a) 0 (b) 3.5 (c) 103.5 (d) 104.5 Solution: (b) For an AP a n = a + (n – 1)d= 3.5+ (101 – 1 )x 0 [by given conditions] ∴ = 3.5 Question 3: The list of numbers – 10, – 6, – 2, 2,… is
WebDec 5, 2024 · The sum of the first n terms of an arithmetic progression (AP) is given by: Sn = n/2 [2a + (n-1)d] where a is the first term of the AP, d is the common difference, and n is … 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 …
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WebIn an AP, if Sn = n (4n+1), then find the AP. Solution 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) ⇒ … optuswebmail in outlook foldersWebNCERT 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 ... portsmouth community classesWebFind the A.P. whose sum to n terms is 2n 2 +n A The required A.P. is 2,6,10,14,... B The required A.P. is 3,7,11,15,... C The required A.P. is 4,8,12,16,... D The required A.P. is 5,10,15,20,... Medium Solution Verified by Toppr Correct option is B) Given, S n=2n 2+n Now, a 1=S 1=2(1) 2+1=3 a 2=S 2−S 1=2(2) 2+2−3=7 optus yes youtubeWebAug 26, 2024 · In an AP, if Sn = n (4n + 1), then find the AP. arithmetic progression class-10 1 Answer +1 vote answered Aug 26, 2024 by Sima02 (49.6k points) selected Aug 26, 2024 … optus yes supportWebCorrect option is A) Given, S n=4n 2−n When n=1 , S 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 , S 2=4(2) 2−1=16=2=14 The sum of the first 2 terms is equal to a 1+a 2 Therefore, a 1+a 2=14 We have shown, above, that a 1=3 Substituting it in a 1+a 2=14, portsmouth community centerWebJan 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 … portsmouth commonwealth attorney staffWebConsider 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 … portsmouth community services board