Asked by Lucas
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)
Attempt at solution:
U5 = 6(U1)
99 = 3(2[U1] + 2d)
Answers
Answered by
Reiny
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 **
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 **
Answered by
Lucas
Thanks for all your answers, I get it now
Answered by
Blossom
Thanks so much I appreciate I now understand
Answered by
Anonymous
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.
Answered by
Anonymous
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.
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