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Sequences and Series: JEE Main Mathematics Question with Solution

Let

\mathrm{A}_{1}, \mathrm{A}_{2}, \mathrm{A}_{3}

be the three A.P. with the same common difference

\mathrm{d}

and having their first terms as

\mathrm{A}, \mathrm{A}+1, \mathrm{A}+2

, respectively. Let

a, b, c

be the

7^{\text{th }}, 9^{\text{th }}, 17^{\text{th }}

terms of

\mathrm{A}_{1}, \mathrm{A}_{2}, \mathrm{A}_{3}

, respectively such that

\left|\begin{array}{ccc}a & 7 & 1 \\ 2 b & 17 & 1 \\ c & 17 & 1\end{array}\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{\mathrm{d}}{12}

, is equal to

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Step-by-step solutionView

Correct answer

The sum of the first 20 terms of the specified arithmetic progression is 495.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Three arithmetic progressions $A_1, A_2, A_3$ have common difference $d$ and first terms $A, A+1, A+2$ respectively. The terms $a, b, c$ are the $7^{\text{th}}, 9^{\text{th}}, 17^{\text{th}}$ terms of $A_1, A_2, A_3$ respectively. Also, $\begin{vmatrix} a & 7 & 1 \\ 2b & 17 & 1 \\ c & 17 & 1 \end{vmatrix} + 70 = 0$ and $a = 29$ .

02goal

Determine the sum of the first 20 terms of an AP with first term $T_1 = c - a - b$ and common difference $D = \frac{d}{12}$ .

03approach

Express $a, b, c$ in terms of $A$ and $d$ : $a = A + 6d$ , $b = (A+1) + 8d = A + 8d + 1$ , $c = (A+2) + 16d = A + 16d + 2$ . Substitute into the determinant equation to find a linear relation between $A$ and $d$ . Combined with $a = A + 6d = 29$ , solve for $A$ and $d$ . Finally, compute $c - a - b$ and the sum $S_{20}$ .

04execute

Substitute $a, b, c$ into the determinant: $\Delta = \begin{vmatrix} A+6d & 7 & 1 \\ 2(A+8d+1) & 17 & 1 \\ A+16d+2 & 17 & 1 \end{vmatrix} = -70$ . Apply row operations $R_2 \to R_2 - R_3$ : $R_2 - R_3 = [2A+16d+2 - (A+16d+2), 17 - 17, 1 - 1] = [A, 0, 0]$ . Expanding along $R_2$ : $\Delta = -A \begin{vmatrix} 7 & 1 \\ 17 & 1 \end{vmatrix} = -A(7 - 17) = 10A$ . Given $\Delta + 70 = 0 \implies 10A + 70 = 0 \implies A = -7$ . Since $a = A + 6d = 29$ , we get $-7 + 6d = 29 \implies 6d = 36 \implies d = 6$ .

05execute

Compute $b$ and $c$ : $b = A + 8d + 1 = -7 + 8(6) + 1 = 42$ . $c = A + 16d + 2 = -7 + 16(6) + 2 = 91$ . First term of the required AP: $T_1 = c - a - b = 91 - 29 - 42 = 20$ . Common difference of the required AP: $D = \frac{d}{12} = \frac{6}{12} = \frac{1}{2}$ . Now compute the sum of the first 20 terms: $S_{20} = \frac{20}{2} \left[2(20) + (20 - 1)\left(\frac{1}{2}\right)\right] = 10 \left[40 + \frac{19}{2}\right] = 10 \left[\frac{99}{2}\right] = 495$ .

✓verify

Verify the determinant directly: $a = 29$ , $2b = 84$ , $c = 91$ . $\begin{vmatrix} 29 & 7 & 1 \\ 84 & 17 & 1 \\ 91 & 17 & 1 \end{vmatrix} = \begin{vmatrix} 29 & 7 & 1 \\ -7 & 0 & 0 \\ 91 & 17 & 1 \end{vmatrix} = -(-7)(7 - 17) = 7(-10) = -70$ . $-70 + 70 = 0$ , which is consistent. The sum $S_{20} = 495$ is correct.

Hints that build this answer step by step

How can the terms $a$ , $b$ , and $c$ be expressed in terms of the initial term $A$ and common difference $d$ ?
$a = A + 6d$ , $b = A + 1 + 8d$ , $c = A + 2 + 16d$
Evaluating the determinant $\Delta = \left|\begin{matrix} a & 7 & 1 \\ 2b & 17 & 1 \\ c & 17 & 1 \end{matrix}\right|$ , what equation in $a, b, c$ does $\Delta + 70 = 0$ yield?
$\Delta = 10(2b - c) = -70 \implies 2b - c = -7$
Using $2b - c = -7$ and $a = 29$ , what are the values of $A$ and $d$ ?
$A = -7$ and $d = 6$
With $A = -7$ and $d = 6$ , what is the sum of the first 20 terms of the AP with first term $A' = c - a - b$ and common difference $d' = \frac{d}{12}$ ?
495

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why apply

R_2 \to R_2 - R_3

rather than standard expansion?

Because rows 2 and 3 both have identical second and third column entries ( $17$ and $1$ ), subtracting them immediately yields two zeros, making expansion a one-line operation.

Answer

The sum of the first 20 terms of the specified arithmetic progression is 495.

Why each option works or fails

Step-by-step solution

given: Three arithmetic progressions $A_1, A_2, A_3$ have common difference $d$ and first terms $A, A+1, A+2$ respectively. The terms $a, b, c$ are the $7^{\text{th}}, 9^{\text{th}}, 17^{\text{th}}$ terms of $A_1, A_2, A_3$ respectively. Also, $\begin{vmatrix} a & 7 & 1 \\ 2b & 17 & 1 \\ c & 17 & 1 \end{vmatrix} + 70 = 0$ and $a = 29$ .
goal: Determine the sum of the first 20 terms of an AP with first term $T_1 = c - a - b$ and common difference $D = \frac{d}{12}$ .
approach: Express $a, b, c$ in terms of $A$ and $d$ : $a = A + 6d$ , $b = (A+1) + 8d = A + 8d + 1$ , $c = (A+2) + 16d = A + 16d + 2$ . Substitute into the determinant equation to find a linear relation between $A$ and $d$ . Combined with $a = A + 6d = 29$ , solve for $A$ and $d$ . Finally, compute $c - a - b$ and the sum $S_{20}$ .
execute: Substitute $a, b, c$ into the determinant: $\Delta = \begin{vmatrix} A+6d & 7 & 1 \\ 2(A+8d+1) & 17 & 1 \\ A+16d+2 & 17 & 1 \end{vmatrix} = -70$ . Apply row operations $R_2 \to R_2 - R_3$ : $R_2 - R_3 = [2A+16d+2 - (A+16d+2), 17 - 17, 1 - 1] = [A, 0, 0]$ . Expanding along $R_2$ : $\Delta = -A \begin{vmatrix} 7 & 1 \\ 17 & 1 \end{vmatrix} = -A(7 - 17) = 10A$ . Given $\Delta + 70 = 0 \implies 10A + 70 = 0 \implies A = -7$ . Since $a = A + 6d = 29$ , we get $-7 + 6d = 29 \implies 6d = 36 \implies d = 6$ .
execute: Compute $b$ and $c$ : $b = A + 8d + 1 = -7 + 8(6) + 1 = 42$ . $c = A + 16d + 2 = -7 + 16(6) + 2 = 91$ . First term of the required AP: $T_1 = c - a - b = 91 - 29 - 42 = 20$ . Common difference of the required AP: $D = \frac{d}{12} = \frac{6}{12} = \frac{1}{2}$ . Now compute the sum of the first 20 terms: $S_{20} = \frac{20}{2} \left[2(20) + (20 - 1)\left(\frac{1}{2}\right)\right] = 10 \left[40 + \frac{19}{2}\right] = 10 \left[\frac{99}{2}\right] = 495$ .
verify: Verify the determinant directly: $a = 29$ , $2b = 84$ , $c = 91$ . $\begin{vmatrix} 29 & 7 & 1 \\ 84 & 17 & 1 \\ 91 & 17 & 1 \end{vmatrix} = \begin{vmatrix} 29 & 7 & 1 \\ -7 & 0 & 0 \\ 91 & 17 & 1 \end{vmatrix} = -(-7)(7 - 17) = 7(-10) = -70$ . $-70 + 70 = 0$ , which is consistent. The sum $S_{20} = 495$ is correct.