Asked by Amanda
The 10th,4th and 1st terms of an A.P are the three consecutive terms of the G.P and the sum of the 1st 6 terms, taking the 1st term to be 4.
Answers
Answered by
mathhelper
10th term --- a + 9d
4th term --- a + 3d
1sterm ---- 4
so
(4+3d)/(4+9d) = 4/(4+3d)
(4+3d)^2 = 4(4+9d)
16 + 24d + 9d^2 = 16 + 36d
9d^2 - 12d = 0
d(9d - 12) = 0
d = 0 or d = 12/9
if d = 0, we could have terms in the AP that would not change, not very interesting, but nevertheless possible,
the terms would be 4, 4, 4, ....
and the sum of the first 6 terms would simply be 24
note that the 10th, the 4th, and the 1st would form a GP, since
4/4 = 4/4
if d = 12/9 = 4/3
sum(6) = 3(8 + 5(4/3) ) = 44
check: the terms of the AP would be
4, 16/3, 20/3, 8, 28/3, 32/3, 12, 40/3, 44/3, 16
is 8/16 = 4/8 ? YES
sum(6) = 4+16/3+ 20/3+ 8+28/3+32/3 = 44
My answer is correct
4th term --- a + 3d
1sterm ---- 4
so
(4+3d)/(4+9d) = 4/(4+3d)
(4+3d)^2 = 4(4+9d)
16 + 24d + 9d^2 = 16 + 36d
9d^2 - 12d = 0
d(9d - 12) = 0
d = 0 or d = 12/9
if d = 0, we could have terms in the AP that would not change, not very interesting, but nevertheless possible,
the terms would be 4, 4, 4, ....
and the sum of the first 6 terms would simply be 24
note that the 10th, the 4th, and the 1st would form a GP, since
4/4 = 4/4
if d = 12/9 = 4/3
sum(6) = 3(8 + 5(4/3) ) = 44
check: the terms of the AP would be
4, 16/3, 20/3, 8, 28/3, 32/3, 12, 40/3, 44/3, 16
is 8/16 = 4/8 ? YES
sum(6) = 4+16/3+ 20/3+ 8+28/3+32/3 = 44
My answer is correct
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.