The 80th term of an arithmetic sequence is twice the 30th term. If the first term of the sequence is 7, what is the 40th term?
Find 'd' the common difference :
From a1 to a30 is 30 terms ( 29 d's)
From 30 to 80 is fifty d's
7 + 29 d = value of 30th term
then for the 80th term (7+29d) + 50 d = 2( 7 + 29 d )
7 + 79 d = 14 + 58 d
-7 = -21 d
d = 1/3 a1 = 7 then a30 = 16 2/3 a80 = 33 1/3 Check !
So 40th term 7 + 39 ( 1/3) = 20
Find 'd' the common difference :
From a1 to a30 is 30 terms ( 29 d's)
From 30 to 80 is fifty d's
7 + 29 d = value of 30th term
then for the 80th term (7+29d) + 50 d = 2( 7 + 29 d )
7 + 79 d = 14 + 58 d
-7 = -21 d
d = 1/3 a1 = 7 then a30 = 16 2/3 a80 = 33 1/3 Check !
So 40th term 7 + 39 ( 1/3) = 20