Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

:) thanx