Question #cd59a

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Answer 1 · 2018-01-11T16:41:28

58th term is #color(blue)(87.5)#

#S_60=color(blue)(2775)#

The sum of an arithmetic series is given by:

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

The nth term is given by:

#a+(n-1)d#

Where #a# is the first term, #d# is the common difference and #n# is the nth term.

From example:

#4/2(2a+(4-1)d)=17# #:.# #4a+6d=17color(white)(88)[ 1 ]#

#8/2(2a+(8-1)d)=58# #:.# #8a+28d=58color(white)(88)[2]#

Solving [ 1 ] and [ 2 ] simultaneously:

Multiply [ 1 ] by #2# and subtract from [ 2 ]:

#8a+28d=58#

#8a +12d=34#

#0+16d=24=>d=3/2#

Substituting in [ 1 ]:

#4a+6(3/2)=17=>a=2#

For the 58th term:

#2+(58-1)(3/2)=color(blue)(87.5)#

Sum of the first 60 terms:

#S_60=60/2[2(2)+(60-1)(3/2)]=color(blue)(2775)#