The sum of a series is given by
S = (n/2)[a1 + a1 + d(n - 1)] where n is the number of terms, a1 is the first term, a1 + d(n - 1) is the last term, and d is the common difference between terms .... so we have
S = (n/2)[ 7 + 7 + 5(n-1) ] > 5000
(n/2) [ 9 + 5n ] > 5000
n[ 9 + 5n ] > 10000
5n^2 + 9n - 10000 > 0 and solving this ... n > 43 → n = 44
So...the last number said is
a44 = [ 7 +5(44 - 1) ] = [ 7 +5(43) ] = 222
The sum of a series is given by
S = (n/2)[a1 + a1 + d(n - 1)] where n is the number of terms, a1 is the first term, a1 + d(n - 1) is the last term, and d is the common difference between terms .... so we have
S = (n/2)[ 7 + 7 + 5(n-1) ] > 5000
(n/2) [ 9 + 5n ] > 5000
n[ 9 + 5n ] > 10000
5n^2 + 9n - 10000 > 0 and solving this ... n > 43 → n = 44
So...the last number said is
a44 = [ 7 +5(44 - 1) ] = [ 7 +5(43) ] = 222