If the sum of 20 consecutive odd integers is 120, then what is the value of the largest integer?

Question

Please and thank you! Your help is greatly appreciated

Answer 1

S(n)=(n/2)[2a+(n-1)d] where  a=1st term, d=2 (common difference)

120=(20/2)[2a+(20-1)2]

120=10[2a+19*2]

120=10[2a+38]

12=2a+38

12-38=2a

-26=2a

a=-13

x(n)=a+d(n-1)

x(20)=-13+2(20-1)

x(20)=-13+2(19)

x(20)=-13+38

x(20)=25, the largest integer

-13,-11,-9,-7,-5,-3,-1,1,3,5,7,9,11,13,15,17,19,21,23,25 are the integers

 

If you are doing SAT math, this is the way to do the problem. However, here is a trick:

the sum is 120 for 20 consecutive odd integers

the mean is 120/20=6

6 is not one of the integers but it is in the middle of the 20 odd integers

therefore, go up 1 to 7 and down 1 to 5, up 2 to 9, down 2 to 3; and seeing that you want the largest and 6 is in the middle, 7 starts the last 10 integers: 7,9,11,13,15,17,19,21,23,25