The first two terms of a geometric progressio | vedclass

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(B) Let the first term be $a$ and the common ratio be $r$.The first two terms are $a$ and $ar$. Their sum is $a + ar = a(1 + r) = 12$  ........  $(i)$

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$-12$

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(B) Let the first term be $a$ and the common ratio be $r$.The first two terms are $a$ and $ar$. Their sum is $a + ar = a(1 + r) = 12$  ........  $(i)$The third and fourth terms are $ar^2$ and $ar^3$. Their sum is $ar^2 + ar^3 = ar^2(1 + r) = 48$  ........  $(ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{ar^2(1 + r)}{a(1 + r)} = \frac{48}{12}$$r^2 = 4$$r = \pm 2$Since the terms of the geometric progression are alternately positive and negative,the common ratio $r$ must be negative. Therefore,$r = -2$.Substituting $r = -2$ into equation $(i)$:$a(1 + (-2)) = 12$$a(-1) = 12$$a = -12$Thus,the first term is $-12$.

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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