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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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