1. Find the sum of the 1st 21 terms of the progression -10, -8, -6 _ _ _ _ _ _
2. Five people are to be arranged in a row for a photograh. How many arangement are there if married couples in a group insist on sitting next to each other?
Tanx alot

Skynumba1:
1. Find the sum of the 1st 21 terms of the progression -10, -8, -6 _ _ _ _ _ _
2. Five people are to be arranged in a row for a photograh. How many arangement are there if married couples in a group insist on sitting next to each other?
Tanx alot

1. Sum of first 21 terms. 21st term = a + 20d where a=-10 and d (common difference) is -8-(-10)=2.

So, 21st term is -10 + 20x2=-10+40=30.
Effectively -10, -8,...,8, 10 will all cancel out!
Sum =12+14+..+30
Sum=(12+30)x5 arranging in pairs. If 12 is the new first term, 30 will be the 10th term, so there are 5 pairs.

Sum = 42x5=210.

There's a straight formula for this, but i can't remember. Hence, first principle.

2. Since we don't know how many married couples there are, this question doesn't have a definite solution.
But if 5 people are to be arranged in a row for a photograph, there are 5! ways to do this.
I.e. you can put mr. A/B/C/D/E on the far left, 5 ways. Once a person occupies that far left, there are only 4 more people you can put next to the person, hence 5! ways = 5 x4x3x2x1=120, let's assume there is one married couple. They constitute one person. Hence there are now (5-1)! ways to seat everyone, 4! ways., i.e. the three people and the one couple. But remember you can interchange the couple wherever they are seated. Hence 2x4! ways.

If there are two couples, 3! ways to seat everyone. Since you can interchange each couple, you have 2x2x3! ways.

goofyone:


1. Sum of first 21 terms. 21st term = a + 20d where a=-10 and d (common difference) is -8-(-10)=2.

So, 21st term is -10 + 20x2=-10+40=30.
Effectively -10, -8,...,8, 10 will all cancel out!
Sum =12+14+..+30
Sum=(12+30)x5 arranging in pairs. If 12 is the new first term, 30 will be the 10th term, so there are 5 pairs.

Sum = 42x5=210.

There's a straight formula for this, but i can't remember. Hence, first principle.

2. Since we don't know how many married couples there are, this question doesn't have a definite solution.
But if 5 people are to be arranged in a row for a photograph, there are 5! ways to do this.
I.e. you can put mr. A/B/C/D/E on the far left, 5 ways. Once a person occupies that far left, there are only 4 more people you can put next to the person, hence 5! ways = 5 x4x3x2x1=120, let's assume there is one married couple. They constitute one person. Hence there are now (5-1)! ways to seat everyone, 4! ways., i.e. the three people and the one couple. But remember you can interchange the couple wherever they are seated. Hence 2x4! ways.

If there are two couples, 3! ways to seat everyone. Since you can interchange each couple, you have 2x2x3! ways.

tanks alot

S = (n/2)(2a+(n-1)d)
=(21/2)(2x-10 + (20x2))=210

S=a+a+d+a+2d+....+a+(n-1)d
S=na + d + 2d + 3d +...+(nd-3d)+(nd-2d)+(nd-d)

S = na + number of nd terms x nd
How do we get the number of nd terms? Mathematical induction tells you it's (n-1)/2.
S = na +(n-1)nd/2 hence the above formula.