In each of the following,$a$ and $d$ for an $A.P.$ are given. Find the $A.P.$ in each case. $a = -12, d = 3$.
(A) The general form of an $A.P.$ is given by $a, a+d, a+2d, a+3d, \ldots$
Given $a = -12$ and $d = 3$.
First term $T_1 = a = -12$.
Second term $T_2 = a + d = -12 + 3 = -9$.
Third term $T_3 = a + 2d = -12 + 2(3) = -12 + 6 = -6$.
Fourth term $T_4 = a + 3d = -12 + 3(3) = -12 + 9 = -3$.
Thus,the $A.P.$ is $-12, -9, -6, -3, \ldots$
The $n^{th}$ term is given by $T_n = a + (n-1)d = -12 + (n-1)3 = -12 + 3n - 3 = 3n - 15$.
Given $a = -12$ and $d = 3$.
First term $T_1 = a = -12$.
Second term $T_2 = a + d = -12 + 3 = -9$.
Third term $T_3 = a + 2d = -12 + 2(3) = -12 + 6 = -6$.
Fourth term $T_4 = a + 3d = -12 + 3(3) = -12 + 9 = -3$.
Thus,the $A.P.$ is $-12, -9, -6, -3, \ldots$
The $n^{th}$ term is given by $T_n = a + (n-1)d = -12 + (n-1)3 = -12 + 3n - 3 = 3n - 15$.
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