How do you find the sum of the arithmetic series #Sigma(5t-3)# from t=19 to 23?

Question

2109 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-06T12:34:45

# sum_(t=19)^(23) (5t-3) =510#

#t_19=5t-3 = 5*19-3=92 , t_20=5t-3 = 5*20-3=97#

#t_23=5t-3 = 5*23-3=112# . First term #a_1=92# , common

difference is #d=97-92=5# , Last term is #a_5=112#

Number of terms is #5# . Mid term is #(92+112)/2 = 204/2=102#

Therefore Sum of #5# terms is #102*5 =510#

#:. sum_(t=19)^(23) (5t-3) =510# [Ans]

Answer 2 · 2018-01-06T17:01:36

510

#sum_(t=19)^23(5t-3)=sum_(t=1)^23(5t-3)-sum_(t=1)^18(5t-3)#

#=[5sum_(t=1)^23t-3sum_(t=1)^23(1)]-[5sum_(t=1)^18t-3sum_(t=1)^18(1)]#

#=5sum_(t=1)^23t-5sum_(t=1)^18t+3sum_(t=1)^18(1)-3sum_(t=1)^23(1)#

Since #sum_(r=1)^nr=n/2(n+1)# and #sum_(r=1)^n1=n# ;

#sum_(t=19)^23(5t-3)=(5*23/2*24)-(5*18/2*19)+(3*18)-(3*23)#

#=1380-855+54-69#

#=510#