In an ap the sum of first n terms is n/2 3n+5
WebThe sum of the first n terms of an AP is (3n^2/2 + 5n/2). Find its nth term and the 25th term. asked Apr 15, 2024 in Arithmetic Progression by Vevek01 (47.4k points) arithmetic progression; class-10; Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. WebMar 12, 2024 · Sum of n terms in an arithmetic progression is given by the formula S = n 2 [ 2 a + ( n − 1) d] in which a = first term, n = number of terms and d = common difference. Let us understand this concept in brief by taking an example. Consider a man putting 100 rupees in his daughter’s piggy bank, in such a way that, He deposits 100 rupees on ...
In an ap the sum of first n terms is n/2 3n+5
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WebIn an A.P the sum of first n terms is 3n^2/2 + 13n/2. Find the 25th term. ← Prev Question Next Question →. 0 votes. 197k views. asked Sep 30, 2024 in Mathematics by Samantha … WebJul 21, 2024 · S_n=n/2(n+1)(2n+3) Given that nth term of a series is T_n=3n^2+2n hence, the sum S_n of given series up to first n terms S_n=\sum_{n=1}^n T_n =\sum_{n=1}^n(3n^2+2n ...
WebApr 6, 2024 · AP 8 10 (1) 3130 (2) 3520 (3) 2790 (4) 1880 Sn 20 Σ Tn ทะ n S = 5 + 11 + 19 + 29 +41 + FL es es Σ = 0 + 5 + 11 + 0=5+6+ T = (5 Solution For sum of first 20 terms of the … WebExample: Find the sum of the first 20 terms of the arithmetic series if a 1 = 5 and a 20 = 62. Solution: Here, a 1 = 5 and a 20 = 62 and n = 20. If we have given the first term and the last …
WebThe sum of the first n terms of an AP is (5n 2 /2 +3n/2). Find its nth term and the 20th term of this AP. arithmetic progression class-10 1 Answer +2 votes answered Sep 14, 2024 by … 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 sum of first term will be the first term So, First term = a = 3 Finding 2nd term Sum of first 2 terms = 4 First term + Second term = 4 3 + a2 = 4 a2 = 4 – 3 a2 = 1 Hence, Common …
WebIf Sn, the sum of first n terms of an A .P. is given by Sn = 5n2 + 3n, then find its nthterm.Sn = 5n2 + 3n from Mathematics Arithmetic Progressions Class 10 Uttarakhand Board Arithmetic Progressions Advertisement Zigya App If S n, the sum of first n terms of an A .P. is given by S n = 5n 2 + 3n, then find its n th term. Sn = 5n2 + 3n 231 Views
WebExample: Find the sum of the first 20 terms of the arithmetic series if a 1 = 5 and a 20 = 62. Solution: Here, a 1 = 5 and a 20 = 62 and n = 20. If we have given the first term and the last term, then. S = n/2 [a + a n] Therefore, S = 20/2 [5 + 20] = 10 25 = 250. So, the sum of the first 20 terms of the AP is 250. green and white cakesWebMar 29, 2024 · Misc 3 Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively . Show that S3 = 3 (S2– S1) We know that Sum of n terms = n/2 [2a + (n – 1)d] i.e. Sn = n/2 [2a + (n – 1)d] where a is the first term & d is the common difference of the A.P. flowers alexandria kyWebMar 31, 2024 · Given: Sum of the first n-terms is 5n²/2 + 3n/2. To find: The 17th term of the A. P. Solution: Step 1. We find the sum of the first 17 terms. Sum of the first 17 terms is = 5 (17)²/2 + 3 (17)/2 = 5 (289)/2 + 51/2 = 1445/2 + 51/2 = 1496/2 = 748. Step 2. We find the sum of the first 16 terms. Sum of the first 16 terms is = 5 (16)²/2 + 3 (16)/2 ... green and white cake ideasflowers alexandria mnWeb(11) Search the sum the first 20 terms of the numerical series in which 3 rl term is 7 also 7 in term is 2 more than three time its 3 rad term. Solution (12) Stylish an arithmetic series, … flowers alexandria nswWebThe sum of n terms of m arithmetical progressions are S 1 , S 2 , S 3 , _____ S m . The first term and common differences are 1, 2, 3,_____ m respectively. Prove that S 1 + S 2 + S 3 + _____ + S m = 4 1 m n (m + 1) (n + 1). flowers alexandria indianaWebWe know that in an AP sum of the first n terms Sn = (n/2) {2a+ (n-1)d} and the nth term=a+ (n-1)d So (3n/2) {2a+ (3n-1)d} = (n/2) [2 {a+ (3n+1–1)d}+ (n-1)d] =>6a+9nd-3d = 2a+6nd+nd-d.. (This is after cancelling n/2 both sides =>4a+2nd-2d = 0 => 2a+nd-d = 0…. (1) The ratio of the first 2n termsis equal to the sum of the next 2n terms => flower sale