Core 1 arithmetic sequences

[peaches_2479]

can someone help me with these questions im really stuck

1.)a polgon has 10 sides. The lengths of the sides, starting with the smallest, form an arithmetic series. The perimeter of the polygon is 675cm and the length of the longest side is twice that of the shortest side. Find for this series

a.) the common difference
b.) the first term

2.)The sum ofthe first two terms of an arithmetic series is 47. The thirtieth term of this series is -62. find

a.) the first term of the series and the common difference
b.) the sum of the first 60 terms of the series.

3.)James decides to save some money for the six-week holiday. He saves 1p on the first day, 2p on the second day and 3p on the third day and so on.

a.)How much will he save at the end of the holiday (42 days)
b.)If he carried on, how long would it be before ha has saved £100

thanks

[krtxmrtz]

can someone help me with these questions im really stuck

1.)a polgon has 10 sides. The lengths of the sides, starting with the smallest, form an arithmetic series. The perimeter of the polygon is 675cm and the length of the longest side is twice that of the shortest side. Find for this series

a.) the common difference
b.) the first term

The length of the sides form an arithmetic series implies:

s_i+1 = s_i + x

where s_i is the length of side i and x is to be determined, so that:

s₁₀ = s₉ + x
s₉ = s₈ + x
...
s₃ = s₂ + x
s₂ = s₁ + x

From this, by recursive back-substitution:

s₁₀ = s₁ + 9x
s₉ = s₁ + 8x
...
s₃ = s₁ + 2x
s₂ = s₁ + x

The perimeter is:

P = s₁ + s₂ + ... + s₉ + s₁₀ = s₁ + (s₁ + x) + (s₁ + 2x) + ... + (s₁ + 8x) + (s₁ + 9x) = 10s₁ + x(1 + 2 + ... + 8 + 9) = 10s₁ + 45x

On the other hand:

s₁₀ = s₁ + 9x
and, because the longest side is twice that of the shortest side:

s₁₀ = 2s₁

Solving these 2 for s₁:

s₁ = 9x

Substituting this result into the equation for the perimeter:

P = 10s₁ + 45x = 90x + 45x = 135x

from which

x = P/135 = 675/135 = 5 and s₁ = 45

[krtxmrtz]

can someone help me with these questions im really stuck
2.)The sum ofthe first two terms of an arithmetic series is 47. The thirtieth term of this series is -62. find

a.) the first term of the series and the common difference
b.) the sum of the first 60 terms of the series.

a₁ + a₂ = 47
a₃₀ = -62

From a_i+1 = a_i + k you get:

a₂ = a₁ + k

and

a₃₀ = a₂₉ + k = a₂₈ + 2k = ... = a₁ + 29k

so that from the first 2 equations above:

a₁ + (a₁ + k) = 47
or
2a₁ + k = 47

and form the second,

a₁ + 29k = -62

Solving this system you get k = -3, and from,

2a₁ + k = 47

you finally find a₁ = (47 - k) / 2 = 25

The sum of the first 60 terms is:

SUM = na₁ + kn(n-1) / 2 = 60*25 - 3*60*59 / 2 = -3810

If you don't know where this formula comes from I can post a simple proof for you.

[krtxmrtz]

3.)James decides to save some money for the six-week holiday. He saves 1p on the first day, 2p on the second day and 3p on the third day and so on.

a.)How much will he save at the end of the holiday (42 days)
b.)If he carried on, how long would it be before ha has saved £100

Amount saved = 1p + 2p + 3p + ... + 42p

You can apply the formula from my previous post,

SUM = na₁ + kn(n-1) / 2

where now

n = 42
a₁ = 1
k = 1

so that the amount saved comes out to be 903p.

I don't know how many p's (pennies?) are in one pound, but assuming it's 100:

SUM = 100*100 = 10000 = na₁ + kn(n-1) / 2 = n +n(n-1)/2

Rearranging and solving:

n = [-1 +/- sqr(1+80000)] / 2

Using the positive solution (the negative one is meaningless),

n = 140.92 approx.

And because the number of days n must be an integer, finally: n = 141