Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.

Given:

Four numbers are in A.P. 

The sum of these numbers is 50 and the greatest number is 4 times the least.

To do:

We have to find the numbers.  

Solution:

Let the first three terms of the AP be $a-3d, a−d, a+d, a+3d$.

According to the problem,

$a-3d+a−d+a+d+a+3d=50\ ......( i)$

$( a+3d)=4(a-3d)\ .....(ii)$

From $(i)$, we get

$4a=50$

$\Rightarrow a=\frac{50}{4}=\frac{25}{2}$

From $(ii)$, we get

$a+3d=4a-12d$

$4a-a=3d+12d$

$3a=15d$

$a=5d\  .....(iii)$

On putting $a=5$ in equation $(iii)$, we get,

$\frac{25}{2}=5d$

$\Rightarrow \frac{5}{2}=d$

$\Rightarrow d=\frac{5}{2}$

This implies,

$a-3d=\frac{25}{2}-3(\frac{5}{2})=\frac{25-15}{2}=\frac{10}{2}=5$

$a-d=\frac{25}{2}-\frac{5}{2}=\frac{25-5}{2}=\frac{20}{2}=10$

$a+d=\frac{25}{2}+\frac{5}{2}=\frac{25+5}{2}=\frac{30}{2}=15$

$a+3d=\frac{25}{2}+3(\frac{5}{2})=\frac{25+15}{2}=\frac{40}{2}=20$

The required A.P is $5, 10, 15, 20,......$

The first four terms of the A.P. are $5, 10, 15$ and $20$.  

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Updated on: 10-Oct-2022

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