Let 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 , respectively. Let a, b, c be the 7^th,9^th,17^th terms of A_1,A_2,A_3 , respective such that | \beginmatrix a & 7 & 1 2b & 17 & 1 c & 17 & 1 \endmatrix | + 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 \fracd12 , is equal to \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Main 2023 · 25 Jan · Shift 1 · Q44 with Solution

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Correct answer: 495

Write the three A.P.s

Let the three A.P.s be:

$A_1: A, A+d, A+2d, \dots$
$A_2: A+1, A+1+d, A+1+2d, \dots$
$A_3: A+2, A+2+d, A+2+2d, \dots$

Given:

$a$ is the $7^{\text{th}}$ term of $A_1$
$b$ is the $9^{\text{th}}$ term of $A_2$
$c$ is the $17^{\text{th}}$ term of $A_3$

So, $a=A+6d,\qquad b=A+1+8d,\qquad c=A+2+16d$

Also given $a=29$ , hence $A+6d=29 \quad \Rightarrow \quad A=29-6d$

Express $b$ and $c$ in terms of $d$

Using $A=29-6d$ : $b=(29-6d)+1+8d=30+2d$ $c=(29-6d)+2+16d=31+10d$

Use the determinant condition

Given

a & 7 & 1\\ 2b & 17 & 1\\ c & 17 & 1 \end{vmatrix}+70=0$$ Substitute $a=29$: $$\begin{vmatrix} 29 & 7 & 1\\ 2b & 17 & 1\\ c & 17 & 1 \end{vmatrix}+70=0$$ Expand along the third column:

\begin{vmatrix} 29 & 7 & 1\ 2b & 17 & 1\ c & 17 & 1 \end{vmatrix} =1\cdot \begin{vmatrix} 2b & 17\ c & 17 \end{vmatrix} -1\cdot \begin{vmatrix} 29 & 7\ c & 17 \end{vmatrix} +1\cdot \begin{vmatrix} 29 & 7\ 2b & 17 \end{vmatrix}

Compute each minor: $$\begin{vmatrix}2b & 17\\ c & 17\end{vmatrix}=34b-17c$$ $$\begin{vmatrix}29 & 7\\ c & 17\end{vmatrix}=493-7c$$ $$\begin{vmatrix}29 & 7\\ 2b & 17\end{vmatrix}=493-14b$$ Therefore, $$D=(34b-17c)-(493-7c)+(493-14b)$$ $$D=20b-10c$$ Given $D+70=0$: $$20b-10c+70=0$$ $$2b-c+7=0$$ $$c=2b+7$$ 4. **Substitute expressions for $b$ and $c$** We have: $$b=30+2d,\qquad c=31+10d$$ Using $c=2b+7$: $$31+10d=2(30+2d)+7$$ $$31+10d=67+4d$$ $$6d=36$$ $$d=6$$ 5. **Find $b$ and $c$** $$b=30+2(6)=42$$ $$c=31+10(6)=91$$ 6. **Form the required A.P.** Its first term is: $$c-a-b=91-29-42=20$$ Its common difference is: $$\frac{d}{12}=\frac{6}{12}=\frac12$$ So the A.P. is: $$20,\ 20.5,\ 21,\dots$$ 7. **Sum of first 20 terms** Using $$S_n=\frac{n}{2}\left[2\ell+(n-1)r\right]$$ with first term $\ell=20$, common difference $r=\frac12$, and $n=20$: $$S_{20}=\frac{20}{2}\left[2(20)+19\cdot\frac12\right]$$ $$S_{20}=10\left[40+\frac{19}{2}\right]$$ $$S_{20}=10\cdot\frac{99}{2}=495$$ 8. **Final answer** $$\boxed{495}$$ The derived answer matches the stored correct answer.

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View written solutionFree

Correct answer: 495

Write the three A.P.s

Let the three A.P.s be:

$A_1: A, A+d, A+2d, \dots$
$A_2: A+1, A+1+d, A+1+2d, \dots$
$A_3: A+2, A+2+d, A+2+2d, \dots$

Given:

$a$ is the $7^{\text{th}}$ term of $A_1$
$b$ is the $9^{\text{th}}$ term of $A_2$
$c$ is the $17^{\text{th}}$ term of $A_3$

So, $a=A+6d,\qquad b=A+1+8d,\qquad c=A+2+16d$

Also given $a=29$ , hence $A+6d=29 \quad \Rightarrow \quad A=29-6d$

Express $b$ and $c$ in terms of $d$

Using $A=29-6d$ : $b=(29-6d)+1+8d=30+2d$ $c=(29-6d)+2+16d=31+10d$

Use the determinant condition

Given

a & 7 & 1\\ 2b & 17 & 1\\ c & 17 & 1 \end{vmatrix}+70=0$$ Substitute $a=29$: $$\begin{vmatrix} 29 & 7 & 1\\ 2b & 17 & 1\\ c & 17 & 1 \end{vmatrix}+70=0$$ Expand along the third column:

Compute each minor: $$\begin{vmatrix}2b & 17\\ c & 17\end{vmatrix}=34b-17c$$ $$\begin{vmatrix}29 & 7\\ c & 17\end{vmatrix}=493-7c$$ $$\begin{vmatrix}29 & 7\\ 2b & 17\end{vmatrix}=493-14b$$ Therefore, $$D=(34b-17c)-(493-7c)+(493-14b)$$ $$D=20b-10c$$ Given $D+70=0$: $$20b-10c+70=0$$ $$2b-c+7=0$$ $$c=2b+7$$ 4. **Substitute expressions for $b$ and $c$** We have: $$b=30+2d,\qquad c=31+10d$$ Using $c=2b+7$: $$31+10d=2(30+2d)+7$$ $$31+10d=67+4d$$ $$6d=36$$ $$d=6$$ 5. **Find $b$ and $c$** $$b=30+2(6)=42$$ $$c=31+10(6)=91$$ 6. **Form the required A.P.** Its first term is: $$c-a-b=91-29-42=20$$ Its common difference is: $$\frac{d}{12}=\frac{6}{12}=\frac12$$ So the A.P. is: $$20,\ 20.5,\ 21,\dots$$ 7. **Sum of first 20 terms** Using $$S_n=\frac{n}{2}\left[2\ell+(n-1)r\right]$$ with first term $\ell=20$, common difference $r=\frac12$, and $n=20$: $$S_{20}=\frac{20}{2}\left[2(20)+19\cdot\frac12\right]$$ $$S_{20}=10\left[40+\frac{19}{2}\right]$$ $$S_{20}=10\cdot\frac{99}{2}=495$$ 8. **Final answer** $$\boxed{495}$$ The derived answer matches the stored correct answer.

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