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8. Sequences and Series
medium
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
A
$-4$
B
$-12$
C
$12$
D
$4$
(AIEEE-2008)
Solution
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
$a=-12$
Standard 11
Mathematics
