AP EAMCET · Maths · Sequences and Series
\(p, x_1, x_2 \ldots, x_n\) and \(q, y_1, y_2, \ldots, y_{\mathrm{n}}\) are two arithmetic progressions with common differences a and \(\mathrm{b}\) respectively. If \(\alpha\) and \(\beta\) are the arithmetic means of \(x_1, x_2, \ldots x_n\), and \(y_1, y_2, \ldots, y_n\) respectively. Then the locus of \(P(\alpha, \beta)\) is
- A \(\mathrm{a}(x-p)=\mathrm{b}(y-q)\)
- B \(\mathrm{b}(x-p)=\mathrm{a}(y-q)\)
- C \(\alpha(x-p)=\beta(y-q)\)
- D \(p(x-\alpha)=q(y-\beta)\)
Answer & Solution
Correct Answer
(B) \(\mathrm{b}(x-p)=\mathrm{a}(y-q)\)
Step-by-step Solution
Detailed explanation
It is given that \(p, x_1, x_2, x_3 \ldots x_n\) and \(q, y_1, y_2, y_3 \ldots y_n\) are in A.P. whose common difference are \(a\) and \(b\) respectively. \[ \begin{aligned} \therefore \quad x_1 & =p+a, x_n=p+n a \\ y_1 & =q+b, y_n=q+n b \end{aligned} \] Also, given \(\alpha\)…
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