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(B) The $n^{	ext{th}}$ term of an $A$.$P$. is given by $T_n = a_1 + (n-1)d$. Given $A_1, A_2, A_3$ have first terms $A, A+1, A+2$ and common differen

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$495$

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(B) The $n^{	ext{th}}$ term of an $A$.$P$. is given by $T_n = a_1 + (n-1)d$. Given $A_1, A_2, A_3$ have first terms $A, A+1, A+2$ and common difference $d$: $a = A + (7-1)d = A + 6d$ $b = (A+1) + (9-1)d = A + 1 + 8d$ $c = (A+2) + (17-1)d = A + 2 + 16d$ Given $a = 29$,so $A + 6d = 29$. The determinant equation is: $\left|\begin{array}{lll} A+6d & 7 & 1 \ 2(A+1+8d) & 17 & 1 \ A+2+16d & 17 & 1\end{array}ight| + 70 = 0$ Subtracting row $2$ from row $3$: $\left|\begin{array}{lll} A+6d & 7 & 1 \ 2A+2+16d & 17 & 1 \ -A & 0 & 0\end{array}ight| + 70 = 0$ Expanding along the third row: $-(-A) 	imes (7 - 17) + 70 = 0 \Rightarrow 10A + 70 = 0 \Rightarrow A = -7$. Since $A + 6d = 29$,we have $-7 + 6d = 29 \Rightarrow 6d = 36 \Rightarrow d = 6$. Now,$a = 29$,$b = -7 + 1 + 8(6) = 42$,$c = -7 + 2 + 16(6) = 91$. First term of new $A$.$P$. is $c - a - b = 91 - 29 - 42 = 20$. Common difference is $\frac{d}{12} = \frac{6}{12} = 0.5$. Sum of first $20$ terms $S_{20} = \frac{20}{2} [2(20) + (20-1)(0.5)] = 10 [40 + 9.5] = 10 [49.5] = 495$.

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