In a set of four numbers the first three are in G.P. and the last three are in A.P. with a common difference $$6$$. If the first number is same as the fourth, the four numbers are

Question

In a set of four numbers the first three are in G.P. and  the last three are in A.P. with a common difference $$6$$.  If the first number is same as the fourth, the four numbers are

Infinity Learn · Accepted Answer

Let the last three numbers in A.P. be b, b $$+$$ $$6$$, b $$+$$ $$12$$ and the first number be a.Hence the four numbers are a, b, b $$+$$ $$6$$, b $$+$$ $$12Given$$,  $$a=b+12$$                                (1)and$$ a, b, b $$+$$ $$6$$ are in G.P. i.e., $$b2$$ $$=$$ a $$(b$$ $$+$$ $$6)or$$       $$b2$$ $$=$$ $$(b$$ $$+$$ $$12) (b$$ $$+$$ $$6)        $$($$∴a $$=$$ b $$+$$ $$12)$$ or  $$18b$$ $$=$$ – $$72$$ \ b $$=$$ –$$4$$,From Eq. (1),a $$=$$ – $$4$$ $$+$$ $$12$$ $$=$$ $$8$$.Hence, the four numbers are $$8$$, –$$4$$, $$2$$ and $$8$$.

In a set of four numbers the first three are in G.P. and the last three are in A.P. with a common difference 6. If the first number is same as the fourth, the four numbers are

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