Question

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

Solution

## Let the four numbers be $$a-3d, a-d, a+d , a+3d$$Now,$$a-3d+a-d+a+d+a+3d=50$$or, $$4a=50\ \Rightarrow \ a=\dfrac {25}{2}$$Let $$d$$ be a positive integer, then greatest number and least number is $$(a-3d)$$According to question, $$a+3d=4(a-3d)$$or, $$a+3d=4a-12d$$or, $$\boxed {a=5d}$$So, $$\dfrac {25}{2}=5\ d. \Rightarrow \ \boxed {d=\dfrac {5}{2}}$$$$\therefore \$$ The number in $$A.P$$ are:$$\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)$$$$=\boxed {5, 10, 15, 20}$$, when least number $$5=4$$ (greatest number)$$=4\times 5 =\boxed {20}$$Mathematics

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