JEE MAIN - Mathematics (2008 - No. 19)
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
- 4
- 12
12
4
Uitleg
As per question,
$$\,\,\,\,\,\,\,\,\,\,\,\,a + ar = 12\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,a{r^2} + a{r^3} = 48\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
$$ \Rightarrow {{a{r^2}\left( {1 + r} \right)} \over {a\left( {1 + r} \right)}} = {{48} \over {12}}$$
$$ \Rightarrow {r^2} = 4, \Rightarrow r = - 2$$
(As terms are $$=+ve$$ and $$-ve$$ alternately)
$$ \Rightarrow a = - 12$$
$$\,\,\,\,\,\,\,\,\,\,\,\,a + ar = 12\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,a{r^2} + a{r^3} = 48\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
$$ \Rightarrow {{a{r^2}\left( {1 + r} \right)} \over {a\left( {1 + r} \right)}} = {{48} \over {12}}$$
$$ \Rightarrow {r^2} = 4, \Rightarrow r = - 2$$
(As terms are $$=+ve$$ and $$-ve$$ alternately)
$$ \Rightarrow a = - 12$$