Progressions Part 2

Since Sravanth has posted a bit long method , let me post a simpler method:

Let the last $3$ terms which are in arithmetic progression with common difference of $6$ be $(a-6) , a , (a+6)$ .

Due to the condition imposed by the problem poser , the $4$ terms are: $(a+6),(a-6),a,(a+6)$ .

Since first $3$ terms are in $GP$ ,

$\dfrac{a-6}{a+6}=\dfrac{a}{a-6} \\ \Rightarrow (a-6)^2=a(a+6) \\ \Rightarrow a^2-12a+36=a^2+6a \\ \Rightarrow 18a=36 \\ \Rightarrow\boxed{a=2}$

$\text{So , the sum of terms }=(a+6)+(a-6)+a+(a+6) \\ = 4a+6 \\ =4(2)+6 \\ =\huge\boxed{14}$

Cheers!

Taking the first statement into consideration, i.e. the first three numbers are in $GP$ , we get the following sequence if b is the first term, $b,br,br^{2},b$

Now, taking the second statement into consideration, i.e, the last three terms are in $AP$ , we get the following sequence if a is the first term, $(a+12),a,(a+6),(a+12)$

Therefore, we can infer that,

$b=(a+12)$ $. . . . . . (i)$

$br=a$ $. . . . . . (ii)$

$br^{2}=(a+6)$ $. . . . . . (iii)$

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Now, using $(i)$ and $(ii)$ we get, $br=a$ $(a+12)r=a$ $r=\dfrac{a}{(a+12)}$ $. . . . . .(iv)$

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Now, substituting $(i)$ and $(iv)$ in $(iii)$ , we get, $br^{2}=(a+6)$ $(a+12)\times \dfrac{a}{a+12} \times \dfrac{a}{a+12}=(a+6)$ $\dfrac{a^{2}}{a+12}=(a+6)$ $a^{2}=(a+6)(a+12)$ $a^{2}=a^{2}+6a+12a+72$ $18a=(-72)$ $a=\boxed{-4}$

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Now, substituting $a=\boxed{-4}$ in the second statement, i.e, $AP$ , we get the following terms, $[(-4)+12],[(-4)],[(-4)+6],[(-4)+12]$ $8,(-4),2,8$

Therefore the sum of the numbers is $8+(-4)+2+8=\boxed{14}$