The first two terms of a geometric progressio | vedclass

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Question

since $a+a r=a(1+r)=12$  ........  $(i)$

And $a r^{2}+a r^{3}=a r^{2}(1+r)=48 \dots$ .......$(ii)$

So, from Eqs. $(i)$ and $(ii)$

$

Vedclass · Accepted Answer

$-12$

Vedclass · Answer

since $a+a r=a(1+r)=12$  ........  $(i)$

And $a r^{2}+a r^{3}=a r^{2}(1+r)=48 \dots$ .......$(ii)$

So, from Eqs. $(i)$ and $(ii)$

$r^{2}=4$

$r=-2$ (terms are alternately +ve and -ve)

On putting the value of $r$ in $E q . (i)$, we get

$a=-12$

The first two terms of a geometric progression add up to $12.$ the sum of the third and the fourth terms is $48.$ If the terms of the geometric progression are alternately positive and negative, then the first term is

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