Let be the three A.P. with the same common difference and having their first terms as A, A + 1, A + 2, respectively. Let be the terms of respectively such that If then the sum of first 20 terms of an A.P. whose first term is and common difference

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Solution

Q.
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 $7^{t h}, 9^{t h}, 17^{t h}$ terms of $A_{1}, A_{2}, A_{3},$ respectively such that $| \begin{matrix} a & 7 & 1 \\ 2 b & 17 & 1 \\ c & 17 & 1 \end{matrix} |$ $+ 70 = 0 .$ If $a = 29,$ then the sum of first 20 terms of an A.P. whose first term is $c - a - b$ and common difference is $\frac{d}{12}$ , is equal to ________ . [2023]

Ans.
(495)

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

$\Rightarrow$ $| \begin{matrix} A + 6 d & 7 & 1 \\ 2 (A + 1 + 8 d) & 17 & 1 \\ A + 2 + 16 d & 17 & 1 \end{matrix} |$ $+ 70 = 0$

$\Rightarrow (A + 6 d) (0) - 7 [(2 A + 2 + 16 d) - (A - 2 - 16 d)] + 1 [17 (2 A + 2 + 16 d - A - 2 - 16 d)] + 70 = 0$

$\Rightarrow - 7 (A) + 1 [17 (A)] + 70 = 0$

$\Rightarrow 10 A + 70 = 0 \Rightarrow A = - 7$

$\Rightarrow A + 6 d = 29 \Rightarrow - 7 + 6 d = 29 \Rightarrow 6 d = 36 \Rightarrow d = 6$

$c = A + 2 + 16 d = - 7 + 2 + 16 (6) = 91$

$b = A + 1 + 8 d = - 7 + 1 + 8 (6) = 42$

$\Rightarrow c - a - b = 91 - 29 - 42 = 20$

$S_{20} = \frac{20}{2} [2 \times 20 + 19 \times \frac{6}{12}] = 495$

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