KCET · Maths · Sequences and Series
If \( a, b, c \) are three consecutive terms of an \( A P \) and \( x, y, z \) are three consecutive terms of a G.P.,
then thevalue of \( X^{b-c}, Y^{c-a}, Z^{a-b} \)
is
- A \( 0 \)
- B \( X Y Z \)
- C \( -1 \)
- D \( 1 \)
Answer & Solution
Correct Answer
(D) \( 1 \)
Step-by-step Solution
Detailed explanation
Given that, \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in A.P. and \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) are in G.P.
Let \(\mathrm{d}\) be the common difference between A.P. consecutive terms. So,
\(b-a=c-b=d, c-a=2 d\)
Now, \(\chi^{b-c} \cdot y^{c-a} \cdot z^{a-b}\)
\(=x^{-d} \cdot y^{2 d} \cdot z^{-d}=(x z)^{-d} \cdot\left(y^{2}\right)^{-d}\)
Since, \(x z=y^{2}\) then \((x z)^{-d}(x z)^{d}=1\)
Let \(\mathrm{d}\) be the common difference between A.P. consecutive terms. So,
\(b-a=c-b=d, c-a=2 d\)
Now, \(\chi^{b-c} \cdot y^{c-a} \cdot z^{a-b}\)
\(=x^{-d} \cdot y^{2 d} \cdot z^{-d}=(x z)^{-d} \cdot\left(y^{2}\right)^{-d}\)
Since, \(x z=y^{2}\) then \((x z)^{-d}(x z)^{d}=1\)
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