Question

The (m+n)th and (m-n)th terms of a G.P. are p and q respectively.Show that the mth terms is underroot pq

Answer 1

ANSWER:

${(m + n)}^{th} = p - - - - - - (1) \\ {(m - n)}^{th} = q - - - - - - (2) \\ since \: they \: are \: in \: gp \\ \\ from \: (1) \\ {(m + n)}^{th} = p \\ {ar}^{m + n - 1} = p - - - - - (3) \\ from(2) \\ {(m - n)}^{th} = q \\ {ar}^{m - n - 1} = q - - - - - (4) \\ (3) \times (4) \\ = > { {a}^{2} r}^{m + n - 1 + m - n - 1} = pq \\ = > { {a}^{2} r}^{2m - 2} = pq \\ = > { ({ar}^{m - 1}) }^{2} = pq \\ = > taking \: square \: root \: on \: both \: sides \\ = > {ar}^{m - 1} = \sqrt{pq } \\ = > {m}^{th} term \: = \sqrt{pq}$

hence proved

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Answer 2

Answer:

Step-by-step explanation: