Question

There are four numbers such that first three of them from an A.P. and the last three form a G.P. The sum of the first and third number is 2 and that of second and fourth is 26. What are these numbers ?

Answer 1

Let the first three terms be $a-d,a,a+d$ where $a$ is the first term and $d$ be the common difference.

According to the question $a-d+a+d=2$

$\Rightarrow 2a=2$ or $a=\frac{2}{2}=1$

Now, let the last three terms be $a,ar,a{r}^2$

Given:$a+a{r}^2=26$

$\Rightarrow a(1+r^2)=26$

$\Rightarrow 1+r^2=\frac{26}{a}=\frac{26}{1}=26$

$\Rightarrow r^2=26-1=25$

$\therefore r=\pm 5$

$\therefore$ Second term $=a=1$

Third term$=ar=5$ or $-5$

Fourth term$=a{r}^2=25$

Thus, by comparing second and third term we get

$d=a_3-a_2$

$d=5-1$ or $-5-1$

$\therefore d=4$ or $-6$

$\therefore$ we know $a_1=a_2-d$

$a_1=1-4$ or $1-(-6)$

$\therefore a_1=-3$ or $7$

Thus there are two sets of numbers

$\{-3,1,5,25\}$ and $\{7,1,-5,25\}$

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