Let A _ 1 , A _ 2 , A _ 3 be the three A.P. with the same common...

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 \left|

\begin{cases}a & 7 & 1 \\ 2b & 17 & 1 \\ c & 17 & 1\end{cases}

\right| + 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

\dfrac{d}{12}

, is equal to _____ .

Correct answer

Step-by-step solution

From the definition of A.P., we have

$a = A + 6d . . . . \left(i\right)$

$b = A + 1 + 8d . . . . \left(ii\right)$

$c = A + 2 + 16d . . . . \left(iii\right)$

Now, we are given

\left| $\begin{cases}a & 7 & 1 \\ 2b & 17 & 1 \\ c & 17 & 1\end{cases}$ \right| + 70 = 0 $$ ⇒ \left| $\begin{cases}A + 6d & 7 & 1 \\ 2A + 2 + 16d & 17 & 1 \\ A + 2 + 16d & 17 & 1\end{cases}$ \right| + 70 = 0 $$ ⇒ \left| $\begin{cases}A + 6d & 7 & 1 \\ 2 + 4d & 3 & - 1 \\ - A & 0 & 0\end{cases}$ \right| + 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 $And,$ \dfrac{d}{12} = \dfrac{6}{12} = \dfrac{1}{2} $So,$ {S}_{20} = \dfrac{20}{2}\left(40 + 19 \times \dfrac{1}{2}\right) = 5 \times 99 = 495

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