in an AP the sum of first n terms is 5n^2/2+3n/2. find its 20th term
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given sum Tn= 5n^2 / 2 + 3n / 2 ----(1)
let ,n= 1,2,3..
put n = 1 in eq(1)
T1= 5(1)^2 / 2 + 3(1) / 2
T1= 5/ 2 + 3 / 2 = 8 / 2 = 4
T1 =4
first term T1 = 4
now put ,n = 2 in (1),we get,
T2 = 5(2)^2 / 2 + 3(2) / 2
T2 = 20 / 2 + 6 / 2
T2 = 13
now put n = 3 in (1), we get
T3 = 45 / 2+ 9 / 2
T3= 27
common difference (d) = (T3 - T2) - ( T2 - T1)
d = (27 - 13) - ( 13 - 4)
d = 14 - 9 = 5
d = 5, and T1= 4 = a
so now, T20 = a+ 19d
T20 = 4 + 19 ( 5)
T20= 4+ 95 = 99
T20 = 99
hence ,the 20th term = 99
★hope it helps ★
let ,n= 1,2,3..
put n = 1 in eq(1)
T1= 5(1)^2 / 2 + 3(1) / 2
T1= 5/ 2 + 3 / 2 = 8 / 2 = 4
T1 =4
first term T1 = 4
now put ,n = 2 in (1),we get,
T2 = 5(2)^2 / 2 + 3(2) / 2
T2 = 20 / 2 + 6 / 2
T2 = 13
now put n = 3 in (1), we get
T3 = 45 / 2+ 9 / 2
T3= 27
common difference (d) = (T3 - T2) - ( T2 - T1)
d = (27 - 13) - ( 13 - 4)
d = 14 - 9 = 5
d = 5, and T1= 4 = a
so now, T20 = a+ 19d
T20 = 4 + 19 ( 5)
T20= 4+ 95 = 99
T20 = 99
hence ,the 20th term = 99
★hope it helps ★
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