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kasiviv002
05-24-2011, 07:45 PM
A Clothing store ordered 3420 sweaters and sold 20 of them at $100 each in the first week. In the second week, the selling price was lowered by$2, and 40 sweaters were sold. In the third week, the selling price was lowered by another $2, and 60 sweaters were sold. If the pattern continued, a) how many weeks did it take to sell all the sweaters? b) what was the selling price in the final week? Subhotosh Khan 05-24-2011, 08:02 PM A Clothing store ordered 3420 sweaters and sold 20 of them at$100 each in the first week. In the second week, the selling price was lowered by $2, and 40 sweaters were sold. In the third week, the selling price was lowered by another$2, and 60 sweaters were sold. If the pattern continued, a) how many weeks did it take to sell all the sweaters? b) what was the selling price in the final week?

DUPLICATE POST:

So what is the pattern of number of sweaters-being-sold each week?

What type of series is this?

What is the equation for summation for this type of series?

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.

kasiviv002
05-24-2011, 08:16 PM
Each week the number of sweaters being sold increases by 20.
week 1 -20 sweaters
week 2- 40 sweaters and so on
This is an arithmetic series.
Sn= 2/n[2a + (n-1)d]
I tried to use the arithmetic series formula by plugging in 3420 for for 'Sn', 20 for 'a' and 20 for 'd'
and im not sure how to solve for 'n' to get the number of weeks

Subhotosh Khan
05-24-2011, 08:26 PM
Each week the number of sweaters being sold increases by 20.
week 1 -20 sweaters
week 2- 40 sweaters and so on
This is an arithmetic series.
Sn= 2/n[2a + (n-1)d] <<<<<< That equation is wrong. It should be:

S_n \ = \ \frac{n}{2}\cdot [2a \ + \ (n-1)\cdot d]

I tried to use the arithmetic series formula by plugging in 3420 for for 'Sn', 20 for 'a' and 20 for 'd'
and im not sure how to solve for 'n' to get the number of weeks

You should get a quadratic equation in 'n' and solve for 'n'.

kasiviv002
05-24-2011, 08:40 PM
Sn= n/2[2a + (n-1)d]
3420=n/2[2(20) + (n-1)(20)]
6840=n[40 + (n-1)(20)]
6840=n[40 + 20n -20]
6840=n[20 + 20n]
6840=20n + 20n^2

im not sure how to solve for 'n' from here on

never mind, i got it... Thank you

Subhotosh Khan
05-25-2011, 02:40 PM
S[sub:12t4foe8]n[/sub:12t4foe8]= n/2[2a + (n-1)d]
3420=n/2[2(20) + (n-1)(20)]
6840=n[40 + (n-1)(20)]
6840=n[40 + 20n -20]
6840=n[20 + 20n]
6840=20n + 20n^2

n[sup:12t4foe8]2[/sup:12t4foe8] + n + 342 = 0 ? (n +19)(n-18) = 0 ? n = 18
im not sure how to solve for 'n' from here on

never mind, i got it... Thank you