Question

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)

Answer 1

Just use your definitions.

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 **

Answer 2

Thanks for all your answers, I get it now

Answer 3

Thanks so much I appreciate I now understand

Answer 4

5. In an geometric progression the term is 432 and the 4th term is 128. Find the common ratio and the sum to infinity of the progression.

Answer 5

5. In an geometric progression the term is 432 and the 4th term is 128. Find the common ratio and the sum to infinity of the progression.