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

2 Answers
Nov 6, 2017

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

Explanation:

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]

Jan 6, 2018

510

Explanation:

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