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In an ap a 1 an 20 sn 399

WebQuestion If an AP has a = 1, a n = 20 and S n = 399, then find the value of n. Solution Compute the value of n: Formulae: The n th term of an AP is a n = a + n - 1 d and the sum of n terms of a series in an AP is S n = n 2 2 a + ( n - 1) d. Substitute a = 1, a n = 20 in the n th term of an AP formula. WebMar 11, 2024 · answered In an Ap, a=1, an=20, sn=399 then find n Advertisement Answer 3 people found it helpful pragyanrana31 Answer: Sn=n/2 (a+l) . . 399=n/2 (1+20) . . 399*2/21=n . . 19*2=n . . N=38 . . Hope this helps you . . Pls mark Brainliest answer! ️ Find Math textbook solutions? Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 NCERT Class 9 …

In an AP if a = 1, an = 20 and Sn = 399, then n is (A) 19 (B) 21 (C) …

WebAnswer : Let the required numbers in A.P.are (a - d), a and (a + d). Sum of these numbers = (a - d) + a + (a + d) = 3a Product of these numbers = (a - d)× a × (a + d)=a (a 2 - d 2) But given, sum = 24 and product = 440 ∴ 3a = 24 a = 8 and a (a 2 - d 2) = 8 (64 - d 2) = 440 [ a = 8] Or, 64 - d 2 = 55 Or, d 2 = 64 - 55 => d 2 = 9 => d = ± 3 WebJan 23, 2024 · In an AP if a = 1, an = 20 and Sn = 399, then n is (A) 19 (B) 21 (C) 38 (D) 42. 2Pi classes. 36.7K subscribers. bins ccv https://lovetreedesign.com

{Latest Edition} NCERT Exemplar Class 10 Maths Solutions Chapter …

WebIn an AP, if a = 1, an = 20 and Sn = 399, then n is equal to (a) 19 (b) 21 (c) 38 (d) 42 Solution: Question 18: The sum of first five multiples of 3 is (a) 45 (b) 55 (c) 65 (d) 75 Solution: (a) The first five multiples of 3 are 3, 6, 9,12 and 15. Here, first term, a = 3, common difference, d = 6-3 = 3 and number of terms, n = 5 WebIn an AP if a = 1, an = 20 and Sn = 399, then n is _____. CBSE English Medium Class 10. Question Papers 939. Textbook Solutions 33590. MCQ Online Mock Tests 12. Important … WebIf an A.P. has a = 1, t n = 20 and s n = 399, then value of n is : A. 20 B. 32 C. 38 D. 40 Answer: Option C Solution (By Examveda Team) S n = 1 2 ( a + l) × n ⇒ 399 = ( 1 + 20) × n 2 ⇒ 399 × 2 = 21 × n ⇒ n = 399 × 2 21 ⇒ n = 19 × 2 ⇒ n = 38 Join The Discussion Related Questions on Progressions How many terms are there in 20, 25, 30 . . . . . . 140? bins ccs

In an AP if a = 1, an = 20 and Sn = 399, then n is ...

Category:In an AP: (i) given a = 8, an = 62. Sn = 210, find ‘n’ and ‘d’.

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In an ap a 1 an 20 sn 399

(2) In an AP if a=1,an =20, and Sn =399,0 then n is equal a=1an.

WebOct 25, 2015 · In an AP, if a=1,an=20 and Sn= 399, then n=? Advertisement Expert-Verified Answer 603 people found it helpful Brainly User since , an = a+ (n-1)d 20 = 1 + (n-1)d 19 = … WebLas protecciones eléctricas tienen un papel relevante en la seguridad y en la adecuada operación de un sistema eléctrico de potencia. Particularmente, en el caso de la generación y el trasporte, por su importancia requieren un eficiente sistema de protecciones que permita garantizar la integridad de sus elementos y la continuidad del servicio eléctrico.

In an ap a 1 an 20 sn 399

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Web98 Likes, 0 Comments - АВТОМОБИЛЬ•ВЫКУП•ОБМЕН•ПРОДАЖА (@avtomarket04) on Instagram: "⚡️Datsun Mi-Do⚡️ Год выпуска ... WebIn an AP if a=1, an =20 and Sn =399, then n is Q. If an AP has a=1, tn=20 and Sn=399, then the value of n is Q. In an AP, if d = - 4 , n = 7 and an = 4, then a is equal to A) 6 B) 7 C) 20 D) …

WebWe can conclude that 10, 20, 30…, 500 is an AP with common difference, d = 10. First term, a = 10. Let the number of terms in this AP = n. Using nth term formula, an = a + (n – 1)d. 500 = 10 + (n – 1)10. 490 = (n – 1)10. n – 1 = 49. n = 50. Sum of an AP, Sn = (n/2) [ a + an], here an is the last term, which is given] = (50/2) ×[10 ... WebFeb 27, 2012 · In an AP if a=1, an= 20 and sn =399 then n is a) 13 b) 38 c) 21 - Maths - Arithmetic Progressions - 1833845 Meritnation.com Class-10 » Maths Arithmetic …

WebFeb 27, 2012 · In an AP if a=1, an= 20 and sn =399 then n is a) 13 b) 38 c) 21 d)19 - Maths - Arithmetic Progressions WebIn an A.P 1 st term is 1 and the last term is 20. The sum of all terms is =399 then n= .... A 42 B 38 C 21 D 19 Medium Solution Verified by Toppr Correct option is B) The is given that, First term (a)=1 Last term (t n)=20 Sum of terms (S n)=399 We know that, t n=a+(n−1)d S n= 2n(2a+(n−1)d) S n= 2n(a+(a+(n−1)d)) ⇒S n= 2n(a+t n) ⇒399= 2n(1+20)

WebSum of n terms of AP is given by: S n = n 2 2 a 1 + n-1 d where a 1 = first term, d = common difference. S 24-4 = 24-4 2 2 × 1 + 24-4-1 13 ∵ d = 13, n = 24, a 1 = 1 S 20 = 20 2 2 + 19 × 13 = 2490. a n-4 = a 1 + n-4-1 d a 24-4 = 1 + 24-4-1 13 ∵ d = 13, n = 24, a 1 = 1 a 20 = 248. Therefore, the ordered pair S n-4, a n-4 is equal to (2490 ...

WebFeb 22, 2013 · In an A.P. , if a = 1, an= 20 and sn =399, then n is? Share with your friends 1 Follow 0 Jai Prakash, added an answer, on 22/2/13 we know from that formula that, Sn = n/2 (a + an) 399 = n/2 (1 + 20) 399*2 = n (21) n = 798/21 = 38 Therefore n = 38 Was this answer helpful? 3 View Full Answer bins cc serverWebBesides giving the explanation of In an A.P. if a=1,An=20 and Sn=399, then find the value of n.?, a detailed solution for In an A.P. if a=1,An=20 and Sn=399, then find the value of n.? … bins cc telegramdaddy please don\u0027t take the girl songWebIf an AP has a=1, t n=20 and S n=399, then the value of n is A 19 B 32 C 38 D 40 Easy Solution Verified by Toppr Correct option is C) Given, a=1 t n=1+(n−1)d=20 ∴(n−1)d=19 S … bins cc 2021WebIn an AP if a = 1, a n = 20 and S n = 399, then n is 38. Explanation: ∵ S n = n 2 [ 2 a + ( n - 1) d] 339 = n 2 [ 2 × 1 + ( n - 1) d] 798 = 2 n + n ( n - 1) d ........ (i) And a n = 20 ⇒ a + ( n - 1) d = … daddy please tie me upWeb20．（本小题满分14分）已知等差数列 an 的公差d 0，它的前n项和为Sn，若 ... 8 Sn . y2. 1的左、右两个顶点分别为A、B．曲21．（本小题满分14分）已知椭圆x 4. 2. 线C是以A、B. P在第一象限且在曲线C上，直线AP与椭圆相交于另一点T． （1）求曲线C的方程； ... bins cell rangeWebIf in an AP a = 1, t n = 20 and Sn = 399 then n is (a) 19 (b) 21 (c) 38 (d) 42 Report 497 Share Follow Answer peter Reputation: 2.6K University of Lagos, Nigeria 30 July 2024 … daddy please meme