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
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 ofn numbers, starting ata
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)|)))