Question

There are four numbers such that the 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

Hint: Choose 1st three terms as a-d, a, a+d where a is the first term and d is the common difference. And the last 3 terms be \[a,ar,a{r^2}\] where r is the common ratio. Then proceed with the problem and try to go through it.

Complete step-by-step answer:
So Let us take the first three numbers as a-d, a, a+d where a is the first term and d is the common difference.
According to the question, the sum of first and third term is 2,
 So, \[a - d + a + d{\text{ }} = {\text{ }}2\]or, \[2a = 2\] or, we have a = 1
Now let the last three terms \[a,ar,a{r^2}\] be where a is the first term among these 3 terms and r is the common ratio.
Given, \[a + a{r^2} = 26\] and we have a=1;
\[
   \Rightarrow a(1 + {r^2}) = 26 \\
   \Rightarrow 1 + {r^2} = 26 \\
   \Rightarrow {r^2} = 25 \\
   \Rightarrow r = \pm 5 \\
 \]
So second term = a= 1; 3rd term=\[ar\]= \[ \pm 5\] ; 4th term=\[a{r^2}\]= 25;
And now, if we check the 3rd term of the AP and the 2nd term of the GP, we can see they are equal.
So, now, we have, a = 1 and r = \[ \pm 5\]and we have to find d,
We get,
\[
  ar = a + d \\
   \Rightarrow \pm 5 = 1 + d \\
   \Rightarrow d = 5 - 1 = 4 \\
  or \\
   \Rightarrow d = - 5 - 1 = - 6 \\
   \Rightarrow d = - 6,4 \\
 \]
So now the 1st term would be a-d = 1- 4 = 3(for +5) and 1-(-6) = 7 (for -5)
Thus we get two sets of numbers {-3,1,5,25} and {7,1,-5,25}.

Note: Try to remember how to write terms in A.P as well as in G.P, then it will be easy to form an equation. In this problem many silly mistakes regarding the plus and the minus sign can be made. Those mistakes should be avoided and be taken care of.