The sum of three numbers in G.P. is $$14$$. If one is added to the first and second numbers and I is subtracted from the third, the new numbers are in A.P. The smallest of them is

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Infinity Learn · Accepted Answer

Let the numbers be a, ar, $$ar2$$. Then,a $$+$$ ar $$+$$ $$ar2$$ $$=$$ $$14$$ (given)----------(1)Now$$,a $$+$$ $$1$$ ar $$+$$ l, $$ar2-$$ $$1$$ are in A.p.⇒$$2(ar+1)=a+1+ar2$$−$$1$$⇒$$2ar+2=a+ar2----2From$$ $$(1) and (2),$$2ar+2=14-aror$$  $$3ar=$$ $$12or$$  $$ar=4$$  $$--------(3)From$$ (1),$$a+4+4r=L4$$ $$a+4r=10$$  $$--------(4)From$$ (3) and (4),$$a+16a=10$$⇒$$a=2$$,$$8Hence$$, the smallest number is $$2$$.

The sum of three numbers in G.P. is 14. If one is added to the first and second numbers and I is subtracted from the third, the new numbers are in A.P. The smallest of them is

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