# Math

In an arithmetic progression, the 5th term is six times the 1st term and the sum of the first six terms is 99. Find the 1st term and the common difference.

Attempt at solution:
U5 = 6(U1)
99 = 3(2[U1] + 2d)

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Isn't the definition of 5th term (a + 4d)
and that of the first term a ?

so a+4d = 6a
4d = 5a
d = 5a/4 -----> **

Also : sum(6) = (6/2)(2a + 5d)
99 = 3(2a + 5(5a/4) )
33 = 2a + 25a/4
times 4
132 = 8a + 25a

continue to solve for a, then sub it back into **

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