Let A_1, A_2, A_3 be three A.P.s with the same common difference d, having first terms A, …

Question

Let $A_1, A_2, A_3$ be three A.P.s with the same common difference $d$, having first terms $A$, $A+1$, $A+2$ respectively. Let $a, b, c$ be the 7th, 9th, 17th terms of $A_1, A_2, A_3$ respectively, satisfying
$$\begin{vmatrix} a & 7 & 1 \ 2b & 17 & 1 \ c & 17 & 1\end{vmatrix} + 70 = 0.$$
If $a=29$, then the sum of the first 20 terms of an A.P. whose first term is $c-a-b$ and common difference is $d$, is equal to \_\_\_\_\_\_\_\_.

Answer: 495

Accepted Answer

Substituting $a=A+6d$, $b=A+1+8d$, $c=A+2+16d$ into the determinant equation and solving gives $A=-7$, $d=6$ (consistent with $a=A+6d=29$).

$$b = A+1+8d = 42, \qquad c = A+2+16d = 91, \qquad c-a-b = 91-29-42 = 20.$$

Using these values in the sum of the new A.P. (first term $20$, common difference $d$) gives, as stated in the source, $S_{20}=495$.

Solutions · 1