Math, asked by guptaananya2005, 3 months ago

EVALUATE the sum in pic

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Sorry to post in pic.

Options are

AP

GP

HP

None of these​

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Answers

Answered by mathdude500
5

Given Question :-

The value of determinant

\rm :\longmapsto\: \begin{gathered}\sf \left | \begin{array}{ccc}x&y&z\\4&3& 2\\x41&y31&z21\end{array}\right | \end{gathered} = 0 \: then \: x,y,z \: are \: in \:

 \green{\large\underline{\sf{Solution-}}}

Given that,

\rm :\longmapsto\:\begin{gathered}\sf \left | \begin{array}{ccc}x&y&z\\4&3& 2\\x41&y31&z21\end{array}\right | \end{gathered} = 0

Now, it can be rewritten as

\rm :\longmapsto\:\begin{gathered}\sf \left | \begin{array}{ccc}x&y&z\\4&3& 2\\100x + 40 + 1&100y + 30 + 1&100z + 20 + 1\end{array}\right | \end{gathered} = 0

\boxed{\tt{ OP \: R_3 \:  \to \: R_ 3- 100R_1}}

and

\boxed{\tt{ OP \: R_3 \:  \to \: R_ 3- 10R_2}}

So we get

\rm :\longmapsto\:\begin{gathered}\sf \left | \begin{array}{ccc}x&y&z\\4&3& 2\\1&1&1\end{array}\right | \end{gathered} = 0

\boxed{\tt{ OP \: C_2 \:  \to \: C_2 - C_1}}

So, we get

\rm :\longmapsto\:\begin{gathered}\sf \left | \begin{array}{ccc}x&y - x&z\\4& - 1& 2\\1&0&1\end{array}\right | \end{gathered} = 0

\boxed{\tt{ OP \: C_3 \:  \to \: C_3 - C_1}}

So, we get

\rm :\longmapsto\:\begin{gathered}\sf \left | \begin{array}{ccc}x&y - x&z - x\\4& - 1&  - 2\\1&0&0\end{array}\right | \end{gathered} = 0

Expanding along Row 3, we get

\rm :\longmapsto\:  - 2(y - x) + 1(z - x) = 0

\rm :\longmapsto\:  - 2y  + 2x+z - x= 0

\rm :\longmapsto\:  - 2y  + x+z = 0

\rm :\longmapsto\:  2y  = x+z

\bf\implies \:x,y,z \: are \: in \: AP

So, option (a) is correct.

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Learn More :-

1. The determinant value remains unaltered if rows and columns are interchanged.

2. The determinant value is 0, if two rows or columns are identical.

3. The determinant value is multiplied by - 1, if successive rows or columns are interchanged.

4. The determinant value remains unaltered if rows or columns are added or subtracted.

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