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 $$\begin{vmatrix} a & 7 & 1 \\ 2b & 17 & 1 \\ c & 17 & 1 \end{vmatrix} + 70 = 0$$. 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 _____.

Three A.P.s $$A_1, A_2, A_3$$ have the same common difference $$d$$ and first terms $$A, A+1, A+2$$ respectively.

• $$a$$ = 7th term of $$A_1$$ = $$A + 6d$$
• $$b$$ = 9th term of $$A_2$$ = $$(A+1) + 8d = A + 1 + 8d$$
• $$c$$ = 17th term of $$A_3$$ = $$(A+2) + 16d = A + 2 + 16d$$

$$\begin{vmatrix} a & 7 & 1 \\ 2b & 17 & 1 \\ c & 17 & 1 \end{vmatrix} + 70 = 0$$

Expanding along the first row:

$$a(17 - 17) - 7(2b - c) + 1(34b - 17c) + 70 = 0$$ $$-14b + 7c + 34b - 17c + 70 = 0$$ $$20b - 10c + 70 = 0 \implies 2b - c = -7$$

Substituting the expressions:

$$2b - c = 2(A + 1 + 8d) - (A + 2 + 16d) = A = -7$$

Given $$a = 29$$:

$$A + 6d = 29 \implies -7 + 6d = 29 \implies d = 6$$

• $$a = -7 + 36 = 29$$
• $$b = -7 + 1 + 48 = 42$$
• $$c = -7 + 2 + 96 = 91$$

First term of new AP: $$c - a - b = 91 - 29 - 42 = 20$$.

Common difference: $$\dfrac{d}{12} = \dfrac{6}{12} = \dfrac{1}{2}$$.

Sum of first 20 terms:

$$S_{20} = \frac{20}{2}\left[2(20) + 19 \times \frac{1}{2}\right] = 10\left[40 + \frac{19}{2}\right] = 10 \times \frac{99}{2} = 495$$

The answer is $$495$$.