if the sum of n terms of an ap is given by SN equals to 3n^ + 4n. determine the ap and the nth term
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3n^ !!!????? Qstn not clear !! But let me solve it using (square) :-
N=1; 3(1)^2 + 4(1) = 7
N=2; 3(2)^2 + 4(2) = 12 + 8 = 20 !
A=7; 20-7 = 13 = d
Therefore T1 = 7
T2 = 20
T3 = 3३
N=1; 3(1)^2 + 4(1) = 7
N=2; 3(2)^2 + 4(2) = 12 + 8 = 20 !
A=7; 20-7 = 13 = d
Therefore T1 = 7
T2 = 20
T3 = 3३
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