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8. Sequences and Series
medium
If ${p^{th}},\;{q^{th}},\;{r^{th}}$ and ${s^{th}}$ terms of an $A.P.$ be in $G.P.$, then $(p - q),\;(q - r),\;(r - s)$ will be in
A
$G.P.$
B
$A.P.$
C
$H.P.$
D
None of these
Solution
(a) If $a$ and $d$ be the first term and common difference of the $A.P.$
Then ${T_p} = a + (p – 1)d,\;$
${T_q} = a + (q – 1)d$
and ${T_r} = a + (r – 1)d$.
If ${T_p},\;{T_q},\;{T_r}$ are in $G.P.$
Then its common ratio $R = \frac{{{T_q}}}{{{T_p}}} = \frac{{{T_r}}}{{{T_q}}} = \frac{{{T_q} – {T_r}}}{{{T_p} – {T_q}}}$
$ = \frac{{[a + (q – 1)d] – [a + (r – 1)d]}}{{[a + (p – 1)d] – [a + (q – 1)d]}} = \frac{{q – r}}{{p – q}}$
Similarly, we can show that $R = \frac{{q – r}}{{p – q}} = \frac{{r – s}}{{q – r}}$
Hence $(p – q),\;(q – r),\;(r – s)$ be in $G.P.$
Standard 11
Mathematics
