Question

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

Answer 1

First, we have to define what the terms we need to solve the problem are.
The given question they were asking to find the first four numbers in the arithmetic progress and the sum is fifty. Also the least number is four times of greatest number, solution as follows, since we need to know about Arithmetic progression. An arithmetic progression can be given by $a,(a + d),(a + 2d),(a + 3d),...$where $a$is the first term and $d$is a common difference.

Complete step by step answer:
Formula to consider for solving these questions ${a_n} = a + (n - 1)d$
Where $d$is the common difference, $a$is the first term, since we know that difference between consecutive terms is constant in any A.P
Since as per the arithmetic numbers the first four terms are, $a,(a + d),(a + 2d),(a + 3d)$and the sum of the first four numbers means we need to add all the four terms, which equal to the $50$.
Hence adding all the four terms we get $a + (a + d) + (a + 2d) + (a + 3d) = 50$(sum of four terms equals fifty). Now we just adding the terms, $4a + 6d = 50$(taking the common terms out and cancels each other we get) $2a + 3d = 25$(the common term two gets eliminated) $(1)$
Since as per the given question, the greatest number of arithmetic is four times the least number in the arithmetic. That means $4(a) = a + 3d$(a is the least and four times, and the greatest term is a plus-three a). hence, we get $4a = a + 3d \Rightarrow 4a - a = 3d$(a will cross into left hand to cancels each other)
$\Rightarrow 4a - a = 3d \Rightarrow a = d$(after canceling the common term three we get the resultant)
Now put an equal to d into equation one we get; $2a + 3d = 25 \Rightarrow 2a + 3a = 25 \Rightarrow 5a = 25$
Now solving on both sides we get; $a = 5$which means also the $d = 5(a = d)$
Hence the first term $a = 5$ is also the common difference $d = 5$thus we get arithmetic numbers the first four terms are, $a,(a + d),(a + 2d),(a + 3d)$and hence substituting we get $5,10,15,20$.

Note:
Since if we add the first four terms, we get $5 + 10 + 15 + 20 = 50$(hence the sum of the first four terms of the arithmetic progress) we can only able to obtain this after finding the first term and the common difference only.