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QJEEM23Ma90b9: JEE Main 2023 Mathematics - Class 12 JEEM23Ma90b9 Solution: Let be the three A.P. with the same common difference d and…

Let A1,A2,A3\mathrm{A_1,A_2,A_3} be the three A.P. with the same common difference d and having their first terms as A,A+1,A+2\mathrm{A,A+1,A+2} , respectively. Let a, b, c be the 7th,9th,17th\mathrm{7^{th},9^{th},17^{th}} terms of A1,A2,A3\mathrm{A_1,A_2,A_3} , respective such that |a712b171c171|+70=0.

If a=29a=29 , then the sum of first 20 terms of an AP whose first term is c−a−bc-a-b and common difference is d12\frac{d}{12} , is equal to ___________.

Step-by-Step Verified Solution

a=A+6da=A+6 d

amp;b=A+8d+1amp;c=A+16d+2amp;∣aamp;7amp;126amp;17amp;1camp;17amp;1∣=−70amp;⇒∣A+6damp;7amp;12A+16d+2amp;17amp;1A+16d+2amp;17amp;1∣=−70amp;R3→R3−R2,R2→R2−R1amp;⇒∣A+6damp;7amp;1A+10d+2amp;10amp;0−Aamp;0amp;0∣=−70\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}

⇒amp;A=−7amp;a=A+6d=29⇒d=6amp;b=−7+48+1=42amp;c=−7+96+2=91amp;c−a−b=91−29−42=20amp; Sum =202[2×20+19×612]=10[40+192]=495\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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