Given a sequence of 4 numbers, the first three of which are in G.P.

Question

Given a sequence of 4 numbers, the first three of which are in G.P. and the last three are in A.P. with common difference six. If the first and last terms of this sequence are equal, then the last term is:

Accepted Answer

Let the numbers be $a,b,c,d$. G.P.: $b^2=ac$. A.P. (last three, common difference 6): $c-b=6$, $d-c=6$. Given $a=d$.

From $c-b=6$ and $a-c=6$ (since $d-c=6$ and $a=d$): $a-b=12\Rightarrow b=a-12$.

Since $2c=b+d=b+a$: $c=\dfrac{a+b}{2}$, and $b^2=ac=a\cdot\dfrac{a+b}{2}\Rightarrow a^2-2b^2+ab=0$.

Substituting $b=a-12$: $a^2-2(a-12)^2+a(a-12)=0\Rightarrow a^2-2a^2+48a-288+a^2-12a=0\Rightarrow36a=288\Rightarrow a=8$.

Since $d=a=8$, the last term is $8$.

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