The sum of the first n terms of an ap is 4n2 -3n find the ap abd hence find 12th teekm
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Given that sum of the first n terms of an ap is
When we substitute 1 for n, we get sum of 1 term = I term.
When we substitute 2 for n, we get sum of 2 terms = a1+a2
From this we get a2.
Since a1 = S1 we get
first term = 4(1)-3 = 1
S2 = 4(4)-3(2) = 10
a2 = S2-S1 = 10-1 =9
d = a2-a1 = 9-1 =8
Arithmetic progression is
1, 9, 17, 25,......
12th term = a12 = a+11d= 1+88 = 89
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