# The 15th term of an arithmetic series is 52, and the sum of the first 15 terms is 405 how do you find the sum of the first 22 terms?

##### 1 Answer

#### Answer:

#### Explanation:

Recall the following:

#1. color(blue)(|bar(ul(color(white)(a/a)"Arithmetic sequence": t_n=a+(n-1)dcolor(white)(a/a)|)))# where:

#t_n=# term number

#a=# first term

#n=# number of terms

#d=# common difference

#2. color(purple)(|bar(ul(color(white)(a/a)"Arithmetic series": S_n=n/2(2a+(n-1)d)color(white)(a/a)|)))# where:

#S_n=# sum of#n# numbers, starting at#a#

#n=# number of terms

#a=# first term

#d=# common difference

**Determining the First Term and Common Difference**

#t_n=a+(n-1)d#

#52=a+(15-1)d#

#color(darkorange)("Equation"color(white)(i)1): 52=a+14d#

#S_n=n/2(2a+(n-1)d)#

#405=15/2(2a+(15-1)d)#

#405=15/2(2a+14d)#

#color(darkorange)("Equation"color(white)(i)2): 405=15a+105d#

#color(teal)15(52)=color(teal)15(a+14d)#

#780=15a+210d#

#color(white)(xxx)780=15a+210d#

#(-(405)=-(15a+105d))/(color(brown)(375=0a+105d))#

#375=0a+105d#

#375=105d#

#color(green)(|bar(ul(color(white)(a/a)d=25/7color(white)(a/a)|)))#

#780=15a+210d#

#780=15a+210(25/7)#

#780=15a+750#

#30=15a#

#color(green)(|bar(ul(color(white)(a/a)a=2color(white)(a/a)|)))#

**Determining the Sum of the First 22 Terms**

#S_n=n/2(2a+(n-1)d)#

#S_22=22/2(2(2)+(22-1)(25/7))#

#S_22=11(4+21(25/7))#

#S_22=11(4+75)#

#color(green)(|bar(ul(color(white)(a/a)S_22=869color(white)(a/a)|)))#