let A 1 , A 2 , A 3 , A 4 ......A n be n Arithmetic means are instead between 20 and 80. i.e., 20 , A 1 , A 2 , A 3 , A 4 ......A n 80. ∴ The Arithmetic Progression consist of ' n + 2 'terms. let 'd' be the common difference. The first term a = 20 The last term L = 80. but the last term is T n + 2 = a + (n + 2 − 1)d = a + (n + 1)d T n + 2 = 80 a + (n + 1)d = 80 20 + (n + 1)d = 80 (n + 1)d = 60 d=60/n+1 Given that the ratio of the first mean to the last mean is 1:3 i.e.,A1/An=1/3
20 + d/20 + nd=1/3 (A n = T n + 1 = a + nd) 20 +60/n+1/20+n.60/n+1 3(n + 4) = (4n + 1) 3n + 12 = 4n + 1 n = 11.
T.kumar
10 Years ago
hi bollu srinivas.. The Arithmetic Progression consist of ' n + 2 'terms.let 'd' be the common difference.The first term a = 20The last term L = 80.but the last term is T n + 2 = a + (n + 2 − 1)d= a + (n + 1)dT n + 2 = 80a + (n + 1)d = 8020 + (n + 1)d = 80(n + 1)d = 60d=60/n+1
SAI SARDAR
10 Years ago
Srinivas, The solution of above question is 11.It is from you can do the problem by the formula of arithematic mean and take ratio.
Ankit Jaiswal
10 Years ago
hi Srinivas, The solution of above question is 11.It is from you can do the problem by the formula of arithematic mean and take ratio.
manmath
10 Years ago
hii
let A 1 , A 2 , A 3 , A 4 ......A n be n Arithmetic means are instead between 20 and 80. i.e., 20 , A 1 , A 2 , A 3 , A 4 ......A n 80. ∴ The Arithmetic Progression consist of ' n + 2...yes or no