if the sum of n terms of an ap be 3n²+n and its common difference is 6 then its first term is
darkcrafter09:
Just put 1 in place of n
The answer is 4
3n²_n=n/2(2a+(n_1). 3(1)²_1=1/2(2a+(1_1)6. 3_1=1/2 2a. a=2
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Sn = 3n² + n
d = 6
If n = 1,
S1 = 3×1²+1
= 4
S1 = a = 4
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