Someone . best answer will be the ! if the average of the 25 terms of a squence of consecutive integers equals the square of the first term, what is the least possible value of the last term?

a25 = 21

Step-by-step explanation:

In accordance with the given data, this is arithmetic sequence.

If we take that the first term is a1, diference d=1, n-th term is an= a1+ (n-1)d  and the sum of the first n terms of the arithmetic sequence is Sn= n/2(a1+an)

Now we don't know the value of the first term but 25-th term is

a25 = a1+(25-1) · 1 = a1+24 => a25=a1+24

The average value of the n-th terms is Sn/n = (n/2(a1+an))/n = 1/2(a1+an)

average value of the 25-th terms is 1/2(a1+a1+24) = 1/2(2a1+24)

That average value is equal to the square of the first term

1/2(2a1+24) = a1∧2   => a1+12 = a1∧2 => a1∧2 - a1 -12 = 0

Now we will factorize this quadratic equation

a1∧2 -4a1 + 3a1 -12 = 0 => (a1∧2 - 4a1) + (3a1 -12) = a1(a1-4) + 3(a1-4) =>

(a1-4) (a1+3) = 0 => a1 - 4 = 0  or a1 + 3 = 0 => a1 = 4 or a1 = -3

because you want the least possible value of the last ( 25-th) term

we choose  a1= -3 and finally we get

a25 = a1+24 = -3 + 24 = 21 => a25 = 21

God with you!!!

i did it myself and i got 4.2 (b) hope its right

step-by-step explanation: