Question

Find four numbers in A.P whose sum is 20 and the sum of whose squares is 120.

Answer 1

Hint: Assume four numbers as $a - 3d,{\text{ }}a - d,{\text{ }}a + d{\text{ and }}a + 3d$. So, by adding all terms we can get the value of one variable. And we will get the value of another variable from the sum of squares of terms.

Complete step-by-step answer:

As we know, the difference between two consecutive terms of A.P is the same.
So, let us assume an A.P.

Let the four numbers in A.P will be,
$ \Rightarrow a - 3d,{\text{ }}a - d,{\text{ }}a + d$and $a + 3d$.

As, common difference of an A.P is the difference of two consecutive numbers.

So, the common difference of the above A.P will be $(a - d) - (a - 3d) = 2d$.

As, given in the question that the sum of four terms of the A.P is 20.

So, $(a - 3d) + (a - d) + (a + d) + (a + 3d) = 20$
So, on solving the above equation. We get,
$ \Rightarrow 4a = 20$
$ \Rightarrow a = 5$ (1)

Now, as given in the question, the sum of squares of 4 numbers of A.P is 120.
$ \Rightarrow $So, ${(a - 3d)^2} + {(a - d)^2} + {(a + d)^2} + {(a + 3d)^2} = 120$

Solving above equation we get,
$ \Rightarrow ({a^2} - 6d + 9{d^2}) + ({a^2} - 2d + {d^2}) + ({a^2} + 2d + {d^2}) + ({a^2} + 6d + 9{d^2}) = 120$

Solving above equation we get,
$ \Rightarrow 4{a^2} + 20{d^2} = 120$

Now, putting the value of a from equation 1 to above equation. We get,
$ \Rightarrow 100 + 20{d^2} = 120$
$ \Rightarrow $Hence, $d = \pm 1$

Now we got the values of a and d. So, we can put the value of a and d to above assumed numbers.

So, four numbers of the A.P will be,

$ \Rightarrow $Hence, the numbers are 2, 4, 6, 8 or 8, 6, 4, 2.

Note: Whenever we came up with this type of problem then to get values of a and d easily. We assume numbers of A.P such that, on applying the given condition, one out of a and d cancels out. And then using the equation of other given conditions. We can find the values of a and d.