Question

if p,q,r are in AP, show that p th ,q th , r th terms of any GP are themselves in GP

Answer 1

$\underline{\bold{Given}}\longrightarrow \\ p \: and \: q \: and \: r \: are \: in \: ap$

$\underline{\bold{To \: prove}}\longrightarrow \\ pth \: qth \: rth \: terms \: of \: any \: gp \: are \: themselves \: in \: gp$

$\underline{\bold{Concept \: used}}\longrightarrow \\ 1) \: if \: a \:and \: b \: and \: c \: are \: in \: ap \\ a + c = 2b \\ 2)nth \: term \: of \: gp = a \: {r}^{n - 1} \\ 3)if \: three \: terms \: are \: in \: gp \\ {b}^{2} = ac$

$\underline{\bold{Solution}}\longrightarrow \\ p \: and \: q \: and \: r \: are \: in \: ap \: so \\ = > p + r = 2q \\ let \: common \: ratio \: be \: x\\ now \\ pth \: term \: of \: gp \: = a {x}^{p - 1} \\ qth \: term \: of \: gp = a {x}^{q - 1} \\ rth \: term \: of \: gp = a {x}^{r - 1} \\ (pth \: term) \: (rth \: term) =( a {x}^{p - 1} ) \: (a {x}^{r - 1}) \\ = {a}^{2} \: {x}^{p - 1 + r - 1} \\ = {a}^{2} \: {x}^{p + r - 2} \\ = {a}^{2} {x}^{2q - 2} \\ = {a}^{2} {x}^{2(q - 1)} \\ = {a}^{2} \: {( {x}^{q - 1}) }^{2} \\ = {(a {x}^{q - 1}) }^{2} \\ = (qth \: term \: of \: gp)^{2} \\ so \: pth \:and \: qth \: and \: rth \: term \:are \: in \: gp$