Question

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

Answer 1

Hint: Arithmetic progression is the sequence in which there is the common difference between any two consecutive terms. Here first of all we will assume four terms in AP and then will follow the given conditions to simplify and get the resultant value.

Complete step-by-step answer:
Let us assume that the four numbers in Arithmetic Progression are –
 $ a - 3d,{\text{ a - d, a + d, a + 3d}} $ .... (A)
Given that the sum of the four numbers in AP is $ 50 $
Place the values –
 $ \Rightarrow \;a - 3d + a - d + a + d + a + 3d = 50 $
Make the pair of like terms in the above equation –
 $ \Rightarrow \underline {\;a + a + a + a} - \underline {3d - d + d + 3d} = 50 $
Simplify the above equation. Remember like terms with equal values and opposite sign cancel each other.
 $ \Rightarrow 4a = 50 $
Make the required “a” the subject. When the term multiplicative at one side moves to another side, then it goes in the denominator.
 $ \Rightarrow a = \dfrac{{50}}{4} $
Find out the common multiples-
 $ \Rightarrow a = \dfrac{{25 \times 2}}{{2 \times 2}} $
Common multiples from the numerator and the denominator cancel each other.
 $ \Rightarrow a = \dfrac{{25}}{2} $ .... (B)
Also, given that the greatest number is four times the least.
 $ \Rightarrow (a + 3d) = 4(a - 3d) $
Simplify the equation –
 $ \Rightarrow a + 3d = 4a - 12d $
Take like terms on one side of the equation –
 $ \Rightarrow 3d + 12d = 4a - a $
Simplify the above equation –
 $ \Rightarrow 15d = 3a $
Take out common multiples from both the sides of the equation.
 $ \Rightarrow (5 \times 3)d = 3a $
Common multiples from both the sides of the equation cancel each other.
 $ \Rightarrow 5d = a $
Place the values from the equation (B)
 $ \Rightarrow 5d = \dfrac{{25}}{2} $
When the term multiplicative at one side is moved to other side then it goes in the denominator.
 $ \Rightarrow d = \dfrac{{25}}{{2 \times 5}} $
Common multiples from the numerator and the denominator cancels each other.
 $ \Rightarrow d = \dfrac{5}{2} $ .... (C)
Place values of the equation (B) and (C) in (A)-
 $ \Rightarrow \left( {\dfrac{{25}}{2} - \dfrac{{15}}{2}} \right),\left( {\dfrac{{25}}{2} - \dfrac{5}{2}} \right),\left( {\dfrac{{25}}{2} + \dfrac{5}{2}} \right),\left( {\dfrac{{25}}{2} + \dfrac{{15}}{2}} \right) $
When denominators are same, we can add or subtract numerators directly –
 $ \Rightarrow 5,10,15,20 $
So, we can observe that greater number is four times the least number, $ 4 \times 5 = 20 $
So, the correct answer is “5,10,15,20”.

Note: Be careful about the signs of the terms while simplification. When the terms are moved from one side to another, Sign of the term is also changed. Positive terms become negative and vice-versa. When you add one positive and one negative, you have to subtract and give a sign of the bigger number.