F5 math problems ( Arithmetic Sequence )

The second term of an arithmetic sequence is nine times the fifth term and the sum of the first eight terms is 56.Find the first term and the common difference.

Let the first term be a
Common difference be d

Second term, T2 = a + d
Fifth term, T5 = a + 4d

So, a + d = 9(a + 4d)
8a = -35d
a = -35d/8

Sum of the first eight terms, S8 = 8/2 [2a + (8 - 1)d]
So, 56 = 4[2(-35d/8) + 7d]
14 = -35d/4 + 7d
-7d/4 = 14
d = -8

a = -35(-8)/8 = 35

So, the first term is 355, and the common difference is -8.

Let the first term and the common difference be a and d respectively.
The first term of the arithmetic sequence is a
The second = a + d
The third = a + 2d
.
The fifth = a + 4d
.
So the eighth = a + 7d

The sum of the first eight terms is 56
Thus,
[a + (a + 7d)] * 8 / 2 = 56
2a + 7d * 8 = 112
2a + 7d = 14 --- (1)
The second term is nine times the fifth term
Thus,
a + d = 9(a + 4d)
a + d = 9a + 36d
8a + 35d = 0 ---(2)

We have two equations...
2a + 7d = 14 --- (1)
8a + 35d = 0 --- (2)
Multiple (1) by 4,
8a + 28d = 56 --- (3)
(3) - (2)
8a + 28d - (8a + 35d) = 56 - 0
28d - 35d = 56
-7d = 56
d = -8
Put d = -8 into (1)
2a + 7(-8) = 14
2a - 56 = 14
2a = 70
a = 35

Therefore, the first term is 35 and the common difference is -8

The nth term of an Arithmetic sequence is a + (n-1)d where a is the initial term
& d is the common difference.

now 2nd term of the sequence is nine times the 5th term.
thus,

a + (2-1)d = 9[a + (5-1)d]
a + d = 9a + 36d
8a + 35d = 0 ...................(1)

The sum of the first eigth terms equal to 56
[a + a + (8-1)d] * 8/2 =56
4(2a + 7d) = 56
8a + 28d =56 ...............(2)

(1) - (2)
we have 7d = -56, hence d = -8

put this result into (1), we have 8a + 35(-8) = 0
8a - 280 = 0
a = 35

Therefore, the initial term is 35 & the common difference is -8.

Let
the firt term = a
the common difference = d
then
a+d = 9(a + 4d)
4(2a + 7d) = 56
so
a = 35 d = -8