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
Mar 27, 2016

Answer:

#S_22=869#

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
#1#. Start by using the arithmetic sequence formula to express the #15^("th")# term mathematically. Label the simplified equation as equation #1#.

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

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

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

#2#. Use the arithmetic series formula to express the sum of the first #15# terms. Label the simplified equation as equation #2#.

#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#

#3#. Use elimination to subtract equation #2# from equation #1#. Start by eliminating the terms with the variable, #a#. However, since equation #1# does not have the same coefficient for #a# as equation #2#, multiply equation #1# by #color(teal)15#.

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

#780=15a+210d#

#4#. Now that equation #1# has the same coefficient for #a# as equation #2#, subtract equation #2# from equation #1#.

#color(white)(xxx)780=15a+210d#
#(-(405)=-(15a+105d))/(color(brown)(375=0a+105d))#

#5#. Solve for #d# in #color(brown)(375=0a+105d)#.

#375=0a+105d#

#375=105d#

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

#6#. Substitute #d=25/7# into either equation #1# or #2# to determine the value of #a#. In this case, we'll use equation #1#.

#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
#1#. Since you now have the values of #a# and #d#, you can use the arithmetic series formula to find the sum of the first #22# terms. Thus, substitute your known values into the formula.

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

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

#2#. Solve for #S_22#.

#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)|)))#