Mathematics Question and Solution
Calculating x, the last term of the series.
a = 7, first term.
d = 10-7 = 13-10 = 3, common difference.
S = 282, sum of the series.
It implies, calculating n, number of terms.
S = ½(n)(2a+d(n-1))
Therefore;
282 = 0.5n(14+3(n-1))
564 = n(11+3n)
3n²+11n-564 = 0
(n+(11/6))² = (564/3)+(11/6)²
(n+(11/6))² = 6889/36
n = -(11/6)±√(6889/36)
n = -(11/6)±(83/6)
It implies;
n ≠ -(11/6)-(83/6)
n = -(11/6)+(83/6)
n = 72/6
n = 12
Calculating x, Las Term.
L = a+d(n-1)
Where L = x, the last term.
It implies;
Notice.
a = 7, d = 3 and n = 12.
Therefore;
x = 7+3(12-1)
x = 7+33
x = 40
