maverickjin8 wrote:
If the sum of the first n positive odd integers is n^2, what is the sum of all odd numbers from 25 to 79?
A) 39^2-11^2
B) 39^2-12^2
C) 40^2-12^2
D) 79^2-22^2
The answer is C but, I don't know how..
Follow posting guidelines (link in my signatures). Any and all analyses should either be put under spoilers or in the next post.As for this question, look below.
You can solve the question in 2 ways:
Method 1: Sum of an arithmetic progression (=a series such as a1 a2 a3 a4 .... such that a1-a2 = a2-a3 = a3-a4 = constant, difference between consecutive terms remains constant.) =
S = \(\frac {n[2a+(n-1)*d]}{2}\), where n = number of terms, a = 1st term, d = constant difference
For the question asked, odd number from 25 to 79 are 25,27,29,31...77,79 and any 2 consecutive terms differ by 2. Thus, n = (79-25)/2+1 = 28, d = 2, a = 25.
Therefore, Sum = S = \(\frac {n[2a+(n-1)*d]}{2}\) = \(\frac {28[2*25+(28-1)*2]}{2}\) = \(28*52\)= \((40-12)*(40+12)\) = \(40^2 - 12^2\) (as \((a+b)*(a-b) = a^2-b^2\)).
Alternately, Method 2, you are given that sum of n odd integers = \(n^2\).
Thus Sum of odd integers from 25 to 79 = Sum of odd integers from 1 to 79 - Sum of odd integers from 1 to 23 (chose 23 as we need 25 in our final calculations).
==> S = \({n_1}^2 - {n_2}^2\), where \(n_1\) = 40 and \(n_2\)= 12 ==> S = \(40^2 - 12 ^2\).
Thus C is the correct answer.