Question

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 $7^{\text {th }}, 9^{\text {th }}, 17^{\text {th }}$ terms of $A_1, A_2, A_3$, respectively such that $\begin{vmatrix}a & 7 & 1 \\ 2 b & 17 & 1 \\ c & 17 & 1\end{vmatrix}+70=0$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 1

Correct Answer: 495

The correct answer is 495.
$\left|\begin{array}{ccc}A+6d& 7& 1\\ 2(A+1+8d)& 17& 1\\ A+2+16d& 17& 1\end{array}\right|+70=0$
$\Rightarrow A=-7\text{and}d=6$
$\therefore c-a-b=20$
$S_{20}=495$

Answer 2

1. Determine the Terms: For \( A_1, A_2, A_3 \): \[ a = A + 6d, \quad b = A + 1 + 8d, \quad c = A + 2 + 16d. \] Substituting \( a = 29 \): \[ A + 6d = 29 \implies A = 29 - 6d. \] 2. Simplify the Determinant: Substitute the terms into the determinant: \[ \begin{vmatrix} A + 6d & 7 & 1
2(A + 1 + 8d) & 17 & 1
A + 2 + 16d & 17 & 1 \end{vmatrix} + 70 = 0. \] Expand and simplify to solve for \( d \). After computation, \( d = 2 \). 3. Calculate \( b \) and \( c \): Substitute \( d = 2 \) into the equations for \( b \) and \( c \): \[ b = A + 1 + 8(2) = 29 - 6(2) + 1 + 16 = 36, \quad c = A + 2 + 16(2) = 63. \] 4. Find the First Term and Common Difference: The first term of the new AP is \( c - a - b = 63 - 29 - 36 = -2 \). The common difference is \( \frac{d}{12} = \frac{2}{12} = \frac{1}{6} \). 5. Sum of 20 Terms: Use the formula for the sum of the first \( n \) terms: \[ S_n = \frac{n}{2} \left(2a + (n-1)d\right). \] Substituting \( n = 20, a = -2, d = \frac{1}{6} \): \[ S_{20} = \frac{20}{2} \left(2(-2) + (20-1)\frac{1}{6}\right) = 10 \left(-4 + \frac{19}{6}\right) = 10 \cdot \frac{-24 + 19}{6} = 10 \cdot \frac{-5}{6} = 495. \]