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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
Let the first term of the geometric progression be a and common ratio be r
Now as per given condition :
$$a+ar= a(1+r) =12$$ (1)
$$ar^2+ar^3 = 48$$ (2)
we get $$ar^2(1+r) =48$$ (3)
Dividing (1) and (3)
we get :
$$1/r^2 = 1/4$$
we get $$1/r = 1/2 or -1/2 $$
but since terms are alternatively positive and negative
we get $$r = -2$$
Substituting
we get $$a(-1) =12 $$
we get $$a=-12$$
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