AS level arithmetic series!!! homework help?

Question

the sum of the first two trms of an arithmetic series is 47.

the thirteenth term of this series is -62. find

i) the first term of the series and the common difference,
ii) the sum of the first 60 terms of the series

Answer 1

i)
a + a+d=2a+d=47
a+12d=-62

Simultaneous equation to solve a and d.
(a is the first term, d is the difference).

ii)
Use:

S = n(a+l)/2
or
S = n(2a + (n-1)d)/2

For n=60

Answer 2

First of all an arithmetic series is the the sum of the arithmetic sequence.

Let the terms of the sequence be u1, u2, u3,...un

Then we have 1st term + 2nd term = 47 or u1 + (u1 +d) = 47.

2U1 + d = 47 Equation (1)

Next use the closed form solution for an arithmetic sequence:

xn = a + (n-1)d (n = 1,2,3...) to give u13 = u1 + (13-1)d

Hence u1 + 12d = -47 Equation (2)

(1) and (2) are simultaneous equations and can be solved to show d = -171/23.

Substituting back into (1) gives:

2u1 + (-171/23) = 47

2u1 = 47 + 171/23

u1 = 626/23

For the sum of the first 60 terms use:

Sn = n/2[2a + (n -1) d]

S60 = 60/2 [2*626/23 + (60 -1) 171/23]

Answer 3

Call the terms in the series a1, a2, a3, etc.

Then a2 = a1 + delta
a3 = a2 + delta = a1 + 2 delta
a4 = a1 + 3 delta
a5 = a1 + 4 delta
etc.

We are given that:
a1 + a2 = 47
a1 + a1 + delta = 47
2 a1 + delta = 47 (equation 1)

also:
a13 = -62
a1 + 12 delta = -62

Which implies that
2 a1 + 24 delta = -124 (equation 2)

Form equation 2 minus equation 1:
2 a1 + 24 delta - 2 a1 - delta = -124 - 47
23 delta = -171
delta = -7.434782609...

The rest follows easily.

The sum of the first 60 terms of an arithmetic series is:
sum = [(a1 + a60) / 2 ] * 60

Answer 4

a1 + (a1+d) = 47
a1 + 12d = -62

a1= -12d-62

2(-12d-62) + d = 47

-23d-124=47
-23d = 171
d = -171/23
a1 = 2052/23 - 62 = 626/23

ii) S=60/2(2*626/23 + 59*(-171/23)) =

= 30(1252/23-10089/23) = 30*(-8837/23) = -265110/23