Question

The sum of four consecutive terms of an A.P. is 20 and the squares of their squares is 120. Find these numbers?

Answer 1

Hint: A.P. means Arithmetic Mean. Add four terms in sequence and then find the product of this consecutive term and solve these two equations to find the series.

Complete step-by-step answer:
Let the four consecutive terms of an A.P.
\[(a-3d),\,(a-d),\,(a+d),\,(a-3d)\]
So according to the question:
\[(a-3d)+(a-d)+\,(a+d)+\,(a+3d)\,=\,20.................(1)\]
As the product of these consecutive terms:
\[(a-3d)\,(a\,-d)\,\,(a+d)\,(a-3d)\,=\,120.................(2)\]
On solving first equation:
\[\begin{align}
  & 4a\,=\,20 \\
 & a\,=\,\dfrac{20}{\begin{align}
  & 4 \\
 & a\,=\,5 \\
\end{align}} \\
\end{align}\]
By putting the value of in equation 2:
\[(a-3d)\,(a-d)\,(a+d)\,(a+3d)\,=\,120\]
\[{{a}^{2}}+9{{d}^{2}}-6ad+{{a}^{2}}+{{d}^{2}}-2ad+{{a}^{2}}+{{d}^{2}}+2ad+{{a}^{2}}+9d{}^{2}+6ad=120\]
\[4{{a}^{2}}+20{{d}^{2}}=120\]
Now put the value of a,
\[\begin{align}
  & 4{{(5)}^{2}}+20{{d}^{2}}=120 \\
 & 100+20{{d}^{2}}=120 \\
 & 20{{d}^{2}}=20 \\
 & {{d}^{2}}=\dfrac{20}{\begin{align}
  & 20 \\
 & d=\sqrt{1} \\
\end{align}} \\
\end{align}\] =
\[d=+1,-1\]
So A.P. is,
\[a-3d,\,a-d,\,a+d,\,a+2d\]
\[2,\,4,\,6,\,8\]
Neglecting the negative value as A.P. cannot be negative.

NOTE: A.P. is a series in which the difference between any two terms should be equal. If the difference between the terms is not equal then this is not an A.P., there are two concepts in A.P. first the number of terms in A.P. and second is sum of terms of an A.P. Both have different formulas to solve. A.P. is a sequence of numbers having equal difference. A.P. can be finite or infinite. In A.P.
A is first term, d is difference between the two terms, n is number is number of terms. the value of the term, is the sum of all terms.
L is the last term .We can also find the term value or number of terms from the last. A.P. is very helpful to find different series values. In A.P. value can increase or decrease also. If the value of d is positive then it will increase and if the value of d is negative then AP will decrease.