Find the 3rd term of the arithmetic series described as : a{1} =102, a{n}= 57 and S{n}=795?

I just need to find the 3rd term. Thats all the question gives me.

my figureing is this, and its prolly wrong: but anyhoow A{1} = 102 + (n-1)d A{n} = 102 + (n-1)d = 57 57-102 = (n-1)d -45 = (n-1)d; for whatever nth term it is ................................................................. Sn = {102( 0d + 1d + 2d + ... + (n-1)d)} +Sn = {102((n-1)d + (n-2)d + (n-3)d + ... + 0d )} ------------------------------------------------ 2 Sn = {204((n-1)d + (n-1)d + (n-1)d + ... + (n-1)d)} 2(795) = 204(-45 + -45 + -45 + -45 + ... + -45) 1590 = 204(-45*n) 1590 1590 -------- = n = - ----- 204*-45 9180 -45 = ((-159/918)-1)d -45 = (-1077/918)d -45(918) -------- = d = 41310/1077; or abt, 38.3565 -1077 .................................................. A{3} = 102 + (41310/1077)(2) =abt. 178.7131 But like I said, thats just my interpretation of it all

\[S_n = \frac{n}{2}(a_1 + a_n)\]\[a_n = a_1 + (n-1)d\]\[S_n = 795\]\[a_n = 57\]\[a_1 = 102\] \[\implies 57 = 102 + (n-1)d\]\[\implies 57-102 = (n-1)d\]\[\implies n = \frac{-45}{d} + 1 \] \[\implies 795 = (\frac{-45}{d}+1)(102 + 57)\]\[\implies 5 = \frac{-45}{d} + 1\]\[\implies d = \frac{-45}{4} = -12\] \[\implies a_3 = a_1 + (3-1)(-12)\]\[\implies a_3 = 102 - 24 = 78\]

Good thinking amistre. Just not quite correct in the execution.

:) thanx

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