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 $r$.Given that the sum of the first two terms is $a + ar = a(1 + r) = 12$  $(i)$.The sum of the

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$-12$

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(B) Let the first term be $a$ and the common ratio be $r$.Given that the sum of the first two terms is $a + ar = a(1 + r) = 12$  $(i)$.The sum of the third and fourth terms is $ar^2 + ar^3 = ar^2(1 + r) = 48$  $(ii)$.Dividing equation $(ii)$ by equation $(i)$,we get:$\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. Thus,$r = -2$.Substituting $r = -2$ into equation $(i)$:$a(1 + (-2)) = 12$$a(-1) = 12$$a = -12$.Therefore,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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