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  4. There are four numbers of which the first three are in G.P. and the last three are in A.P., whose common difference is 6. If the first and the last numbers are equal then two other numbers are

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Q. There are four numbers of which the first three are in G.P. and the last three are in A.P., whose common difference is $6$. If the first and the last numbers are equal then two other numbers are

Sequences and Series

Solution:

Let the last three numbers in A.P. be $a, a+6, a+12$,
then the first term is also $a+12$.
But $a+12, a, a+6$ are in G.P.
$\therefore a^{2}=(a+12)(a+6)$
$ \Rightarrow a^{2}=a^{2}+18 a+72$
$\therefore a=-4 .$
$\therefore $ The numbers are $8,-4,2,8 $