Question

The sum of 28 consecutive odd positive integers is a perfect cube. Determine the smallest possible set? What is the first and last terms in the sequence. Plz helppp!

Answer 1

Answer:

71 + 73 + ..... + 125 = 2,744 = $14^{3}$.

Step-by-step explanation:

The smallest set starts at 71.  The last term is 125.

The total is 2,744, which is $14^{3}$.

Answer 2

The smallest possible set is 71,73,75,....125. The first and last terms in the sequence is 71 and 125 respectively.

What is airthmatic sequence?

Geometric sequence is the sequence in which the next term is obtained by adding or substracting the previous term with the same number for the whole series.

The sumation of airthmatic sequence is find out using the following formula.

$S=\dfrac{n}{2}(a+l)$

Here, (a) is the first term of the sequence, (l) is the last term and (n) is the number of terms in sequence.

Let x is the first and the smallest term of the set.  This set consists 28 consecutive odd positive integers. Thus, the sum of this set will be,

$\sum=x+(x+2)+(x+4)+...+(x+54)\\\sum=28x+(2+4+6+....+54)\\\sum=28x+[\dfrac{27}{2}(2+54)]\\\sum=28x+756\\\sum=28(x+27)\\\sum=2\times2\times7(x+27)$

Now here 7 is prime number. The value of (x+27) should be perfect square of 7 and 2 to make it perfect cube. thus,

$x+27=7^2\times2\\x=98-27\\x=71$

Put this value to find the sequence as,

$71,71+2,71+4......,71+34\\71,73,75,.......125$

The first term in this set is 71 and last term is 125.

Hence, the smallest possible set is 71,73,75,....125. The first and last terms in the sequence is 71 and 125 respectively.

Learn more about the airthmatic sequence here;

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