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8. Sequences and Series
medium
If $(y - x),\,\,2(y - a)$ and $(y - z)$ are in $H.P.$, then $x - a,$ $y - a,$ $z - a$ are in
A
$A.P.$
B
$G.P.$
C
$H.P.$
D
None of these
Solution
(b) $(y – x),\,2(y – a),(y – z)$ are in $H.P.$
==> $\frac{1}{{y – x}},\frac{1}{{2(y – a)}},\frac{1}{{y – z}}$ are in $A.P.$
==> $\frac{1}{{2(y – a)}} – \frac{1}{{(y – x)}} = \frac{1}{{y – z}} – \frac{1}{{2(y – a)}}$
==> $\frac{{y – x – 2y + 2a}}{{(y – x)}} = \frac{{2y – 2a – y + z}}{{(y – a) – (z – a)}}$
$ \Rightarrow \frac{{ – x – y + 2a}}{{(y – x)}} = \frac{{y + z – 2a}}{{(y – z)}}$
$ \Rightarrow \frac{{(x – a) + (y – a)}}{{(x – a) – (y – a)}} = \frac{{(y – a) + (z – a)}}{{(y – a) – (z – a)}}$
==> $\frac{{(x – a)}}{{(y – a)}} = \frac{{(y – a)}}{{(z – a)}}$
$(x – a),(y – a),(z – a)$ are in $G.P.$
Standard 11
Mathematics
