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The sum of three numbers in A.P. is $30$ and the ratio of the first number to the third the number is $3:7$. Find the numbers.
Given: The sum of three numbers in A.P. is $30$ and the ratio of the first number to the third the number is $3:7$.
To do: To find the numbers.
Solution:
It is given that, the sum of three numbers in A.P. $=30$
The ratio of the first number to the third number is $3: 7$
Let us assume the $3$ numbers which are in A.P. are, $a−d,\ a,\ a+d$
Now adding $3$ numbers $=a−d+a+a+d=30$
$\Rightarrow 3a=30$
$\Rightarrow a=\frac{30}{3}$
$\Rightarrow a=10$
Given ratio $3:7=a−d:a+d$
$\Rightarrow \frac{3}{7}=\frac{( a−d)}{( a+d)}$
$\Rightarrow ( a−d)7=3( a+d)$
$\Rightarrow 7a−7d=3a+3d$
$\Rightarrow 7a−3a=7d+3d$
$\Rightarrow 4a=10d$
$\Rightarrow 4( 10)=10d$
$\Rightarrow 40=10d$
$\Rightarrow d=\frac{40}{10}$
$\Rightarrow d=4$
Therefore, the numbers are $a−d=10−4=6$
$a=10$
$a+d=10+4=14$
$\therefore 6,\ 10,\ 14,\ ...... $ are in A.P.
Updated on: 10-Oct-2022
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