The sum of the first six term of an arithmetic progression is 21 and the 7th term is three time the sum of the third and fourth term find the first term and the common difference.

3 answers

The sum of the first six term of an arithmetic progression is 21
---- > (6/2)(2a + 5d) = 21
2a + 5d = 7

the 7th term is three time the sum of the third and fourth term
---> a+6d = 3(a+2d + a+3d)
a+6d = 6a + 15d
-5a = 9d
a = -9d/5
sub into 2a + 5d = 7
2(-9d/5) + 5d = 7
times 5
-18d + 25d = 35
7d = 35
d = 5, then a = -9(5)/5 = -9

check:
sum(6) = 3(-18+25) = 21
sum of 3rd and 4th = a + 2d + a + 3d = 2a + 5d = -18 + 25 = 7
7th term = a + 6d = -9 + 30 = 21 , which is 3 times 7
all is ok

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