JEE MAIN - Mathematics (2023 - 25th January Morning Shift - No. 21)
Let $$\mathrm{A_1,A_2,A_3}$$ be the three A.P. with the same common difference d and having their first terms as $$\mathrm{A,A+1,A+2}$$, respectively. Let a, b, c be the $$\mathrm{7^{th},9^{th},17^{th}}$$ terms of $$\mathrm{A_1,A_2,A_3}$$, respective such that $$\left| {\matrix{ a & 7 & 1 \cr {2b} & {17} & 1 \cr c & {17} & 1 \cr } } \right| + 70 = 0$$.
If $$a=29$$, then the sum of first 20 terms of an AP whose first term is $$c-a-b$$ and common difference is $$\frac{d}{12}$$, is equal to ___________.
Answer
495
Explanation
$a=A+6 d$
$$ \begin{aligned} & b=A+8 d+1 \\\\ & c=A+16 d+2 \\\\ & \left|\begin{array}{ccc} a & 7 & 1 \\ 26 & 17 & 1 \\ c & 17 & 1 \end{array}\right|=-70 \\\\ & \Rightarrow\left|\begin{array}{ccc} A+6 d & 7 & 1 \\ 2 A+16 d+2 & 17 & 1 \\ A+16 d+2 & 17 & 1 \end{array}\right|=-70 \\\\ & R_{3} \rightarrow R_{3}-R_{2}, \quad R_{2} \rightarrow R_{2}-R_{1} \\\\ & \Rightarrow\left|\begin{array}{ccc} A+6 d & 7 & 1 \\ A+10 d+2 & 10 & 0 \\ -A & 0 & 0 \end{array}\right|=-70 \end{aligned} $$
$$ \begin{aligned} \Rightarrow \quad & A=-7 \\\\ & a=A+6 d=29 \Rightarrow d=6 \\\\ & b=-7+48+1=42 \\\\ & c=-7+96+2=91 \\\\ & c-a-b=91-29-42=20 \\\\ & \text { Sum }=\frac{20}{2}\left[2 \times 20+19 \times \frac{6}{12}\right]=10\left[40+\frac{19}{2}\right]=495 \end{aligned} $$
$$ \begin{aligned} & b=A+8 d+1 \\\\ & c=A+16 d+2 \\\\ & \left|\begin{array}{ccc} a & 7 & 1 \\ 26 & 17 & 1 \\ c & 17 & 1 \end{array}\right|=-70 \\\\ & \Rightarrow\left|\begin{array}{ccc} A+6 d & 7 & 1 \\ 2 A+16 d+2 & 17 & 1 \\ A+16 d+2 & 17 & 1 \end{array}\right|=-70 \\\\ & R_{3} \rightarrow R_{3}-R_{2}, \quad R_{2} \rightarrow R_{2}-R_{1} \\\\ & \Rightarrow\left|\begin{array}{ccc} A+6 d & 7 & 1 \\ A+10 d+2 & 10 & 0 \\ -A & 0 & 0 \end{array}\right|=-70 \end{aligned} $$
$$ \begin{aligned} \Rightarrow \quad & A=-7 \\\\ & a=A+6 d=29 \Rightarrow d=6 \\\\ & b=-7+48+1=42 \\\\ & c=-7+96+2=91 \\\\ & c-a-b=91-29-42=20 \\\\ & \text { Sum }=\frac{20}{2}\left[2 \times 20+19 \times \frac{6}{12}\right]=10\left[40+\frac{19}{2}\right]=495 \end{aligned} $$
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