WebAug 27, 2024 · The nth term of an Arithmetic progression is 4 . Common difference of the Arithmetic progression is 2 . The sum of the n terms of the Arithmetic progression is - 14 . This implies ; Using the formula , to find the nth term of the AP ! = a + ( n - 1 ) d }= 4 d = 2 4 = a + ( n - 1 )24 = a + 2n - 2 4 + 2 = a +2n 6 = a + 2n a + 2n = 6 equation−1 WebMar 22, 2024 · Given a = 8, an = 62, Sn = 210, Since there are n terms, 𝑙 = an = 62 We use the formula Sn = 𝒏/𝟐 (𝒂+𝒍) Putting a = 8, Sn = 210, 𝑙 = an = 62 210 = 𝑛/2 (8+62) 210 × 2=𝑛 × (70) 420 = …
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WebMar 29, 2024 · And, the formula to calculate sum of first n terms of an AP, that is S n, is given by, S n = n 2 ( 2 a + ( n − 1) d) ⋯ ⋯ ( i i) Now, In the given question we have, a n = 4, d = 2. Using these values in equation ( i) , we can write, a + ( n − 1) 2 = 4. Applying distributive law, we get, a + 2 n − 2 = 4. WebMar 23, 2024 · In an AP given an=4 d=2 sn=-14 find n and a Class 10 Maths Chapter 5 Exercise 5.3 Question 3 ka 8 425 views Premiered Mar 23, 2024 15 Dislike Share Save Garg Tutorials 12.4K... dan frazier is back: the allied signets
Sum of N terms of an AP - Formula, Examples Sum of …
Web(viii) Given an=4, d=2, Sn=−14, find n and a. (ix) Given a=3,n=8,S=192, find d. (x) Given l=28,S=144, and there are total 9 terms. Find a. Q. In an AP (i) Given a =5,d=3,an=50, find n and Sn. (ii) Given a=7,a13=35, find d and S13. (iii) Given a12=37,d=3, find a and S12. (iv) Given a3 =15,S10 =125, find d and a10. (v) Given d=5,S9=75, find a and a9. WebIn an AP, given a n=4,d=2,S n=−14, find n and a. Medium View solution > View more More From Chapter Sequences and series View chapter > Revise with Concepts Introduction to Arithmetic Progression - 2 Example Definitions Formulaes Arithmetic Mean Example Definitions Formulaes Learn with Videos General Form of an Arithmetic Progression 8 6 WebHere, a n = 4, d = 2 S n = –14 We know that, a n = a + (n -1)d 4 = a + (n -1) x 2 4 = a + 2n - 2 a + 2n = 6 a = 6 - 2 n ....(i) And. S n = [2a+ (n - 1)d] - 14 ... dan frazier texas department of banking