the sum of first 8 terms of an arithmetic progression is 156.the ratio of its 12th term to its 68th term is 1:5.calculate the first term and the fifteenth term.
please help this questions please.
5 answers
-
sum of 8 terms is 156
(8/2)(2a + 7d) = 156
4(2a + 7d) = 156
2a + 7d = 39
(a+11d) : (a + 67d) = 1 : 5
(a+11d)/(a+67d) = 1/5
5a + 55d = a + 67d
4a = 12d
a = 3d
sub that into 2a+7d=39
6d + 7d = 39
13d=39
d=3 , then in a = 3d ---> a = 9
first term is 9
term 15 = a+14d = 9+14(3) = 51
check:
sum(8) = (8/2)(18 + 7(3)) = 4(39) = 156 , checks !
term12 = a+11d = 42
term 68 = a+67d = 210
and 42/210 = 1/5 , YEahh!
-
Better explain
-
I haven't get your response yet please help me
-
-
How many terms has the A.P Whose first term is 15 and the last is 57 given that the common difference is 3
-
Maths
