Question

The first term of GP exceeds the second term by 1/2 and the sum to
infinity is 2. Find the GP.​

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: find : \\ sequence \: of \: GP \\ \\ so \: for \: certain \: GP \\ let \: its \: \\ first \: term \: be \: a \: and \: that \: of \: second \\ term \: be \: \: ar \\ \\ according \: to \: first \: condition \\ a = ar + \frac{1}{2} \\ \\ a - \frac{1}{2} = ar \\ \\ar = \frac{2a - 1}{2} \\ \\ r = \frac{2a - 1}{2a} \: \: \: \: \: \: \: \: (1) \\ \\ according \: to \: second \: condition \\ we \: know \: that \\ \: sum \: to \: infinity \: of \: a \: GP \\ is \: given \: by \\ \\ Sn = \frac{a}{1 - r} \\ \\ 2 = \frac{a}{1 - r} \\ \\ = \frac{a}{1 - (\frac{2a - 1}{2a} \: )} \\ \\ = \frac{ \frac{a}{2a - 2a + 1} }{2a} \\ \\ = \frac{ \frac{a}{1} }{2a} \\ \\ 2 = a.2a \\ \\ 2a {}^{2} = 2 \\ \\ a {}^{2} = 1 \\ \\ a = 1 \: \: or \: \: a = - 1$

$if \: we \: take \: a = 1 \\ then \: \: we \: get \\ \\ r = \frac{2a - 1}{2a} \: \: \: \: \: \: \: from \: (1) \\ \\ = \frac{2(1) - 1}{2(1)} \\ \\ = \frac{2 - 1}{2} \\ \\r = \frac{1}{2}$

$if \: we \: take \: a = - 1 \\ then \\ \\ r = \frac{2( - 1) - 1}{2( - 1)} \\ \\ = \frac{ - 2 - 1}{ - 2} \\ \\ = \frac{ - 3}{ - 2} \\ \\ r = \frac{3}{2}$

$but \: since \: \: as \: here \\ sum \: to \: infinity \: of \: GP \: exists \\ |r| < 1 \\ \\ if \: we \: take \: \\ r = \frac{3}{2} \\ \\ \frac{3}{2} = 1.5 > 1 \\ \\ hence \: here \\ r = 1.5 \: \: is \: absurd$

$thus \: then \\ required \: GP \: is \: \\ 1, \frac{1}{2} \: , \: and \: so \: on$