Question

⇒ In a set of 4 numbers , the first three are in G.P and the last three are in A.P with a common difference of 6. If the firs number is same as the fourth , find the four numbers.
⇔ The 4 numbers are 8 , -4 , 2 , 8

Answer 1

_____________________

✎ Question :-

In a set of 4 numbers , the first three are in ɢ.ᴘ and the last three are in ᴀ.ᴘ with a common difference of 6. If the first number is same as the fourth , find the four numbers.

✎ Solution :-

Refer the attachments for the solution !!

Due to some technical issues I had to put it on the paper !!

Hope uh understand !!

_____________________

Answer 2

Given Question :-

• In a set of 4 numbers , the first three are in G.P and the last three are in A.P with a common difference of 6. If the first number is same as the fourth , find the four numbers.

Answer :-

Given :-

• Set of four numbers.
• First three numbers are in GP
• Last three numbers are in AP with common difference 6.
• First number is same as fourth number.

To find :-

• The four numbers

Solution :-

☆ Let the numbers be a, b, c, d.

☆ Since, b, c, d are in AP with common difference 6

$\tt\implies \:c - b = d - c = 6$

$\tt\implies \:c \: = b + 6 \: and \: d \: = c + 6$

$\tt\implies \:c \: = b + 6 \: and \: d \: = b \: + 12$

☆ Also, it is given that first term and fourth term are same

$\tt\implies \:a \: = \: d \: = \: b \: + 12$

☆ So required four numbers are now

$\tt \:  b + 12 \: , \:b \: , \: b + 6, \: b \: + 12$

$\begin{gathered}\bf\red{Now,}\end{gathered}$

$\begin{gathered}\bf\red{According \: to \: statement}\end{gathered}$

$\tt \:  \longrightarrow \: b + 12, \: b, \: b \: + 12 \: are \: in \: GP$

$\tt \:  \longrightarrow \: \dfrac{b}{b \: + 12} = \dfrac{b \: + \: 6}{b}$

$\tt \:  \longrightarrow \: {b }^{2} = (b + 6)(b + 12)$

$\tt \:  \longrightarrow \: {b}^{2} = {b}^{2} + 6b + 12b + 72$

$\tt \:  \longrightarrow \: 18b \: + \: 72 = 0$

$\tt \:  \longrightarrow \: 18b \: = \: - \: 72$

$\tt \:  \longrightarrow \: b \: = \: - \: 4$

$\begin{gathered}\begin{gathered}\bf So \: numbers \: are \: = \begin{cases} &\sf{b \: + 12 = - 4 + 12 = 8} \\ &\sf{b = - \: 4} \\ &\sf{b \: + 6 = - 4 + 6 = 2} \\ &\sf{b \: + \: 12 = - 4 + 12 = 8}\end{cases}\end{gathered}\end{gathered}$