Let $A_1$, $A_2$, $A_3$ be the three A.P. with the same common difference $d$ and having their first terms…
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 $A_1$, $A_2$, $A_3$, respectively such that $\begin{aligned} \begin{vmatrix} a & 7 & 1 \\ 2b & 17 & 1 \\ c & 17 & 1 \end{vmatrix} + 70 = 0 \end{aligned}$. 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 _____ .
Solution
From the definition of A.P., we have
$\begin{aligned}
a &= A + 6d \quad \text{(i)} \\
b &= A + 1 + 8d \quad \text{(ii)} \\
c &= A + 2 + 16d \quad \text{(iii)}
\end{aligned}$
Now, we are given
$
\begin{vmatrix} a & 7 & 1 \\ 2b & 17 & 1 \\ c & 17 & 1 \end{vmatrix} + 70 = 0
\begin{vmatrix} A + 6d & 7 & 1 \\ 2A + 2 + 16d & 17 & 1 \\ A + 2 + 16d & 17 & 1 \end{vmatrix} + 70 = 0
\begin{vmatrix} A + 6d & 7 & 1 \\ 2 + 4d & 3 & -1 \\ -A & 0 & 0 \end{vmatrix} + 70 = 0
$
10A + 70 = 0
$
A = -7
So,
a = A + 6d
29 = -7 + 6d
d = 6
c - a - b = 1 + 2d - A = 13 + 7 = 20
$
And,
$\frac{d}{12}$ = $\frac{6}{12}$ = $\frac{1}{2}$
So,
$
S_{20} = \frac{20}{2} \left(40 + 19 \times \frac{1}{2}\right) = 5 \times 99 = 495
$