Nd the 52nd term and the term named in the problem 40, 140, 240, 340, ...Find a 32
1 answer:
Answer:
5140 and 3140
Step-by-step explanation:
Note there is a common difference d between consecutive terms in the sequence, that is
d = 140 - 40 = 240 - 140 = 340 - 240 = 100
This indicates the sequence is arithmetic with n th term
= a₁ + (n - 1)d
where a₁ is the first term and d the common difference
Here a₁ = 40 and d = 100, thus
= 40 + (51 × 100) = 40 + 5100 = 5140
= 40 + (31 × 100) = 40 + 3100 = 3140
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Answer:
There are 35 different sets of 3 books Joe could choose
Step-by-step explanation:
* Lets explain how to solve the problem
- Combination is a collection of the objects where the order doesn't
matter
- The formula for the number of possible combinations of r objects from
a set of n objects is nCr = n!/r!(n-r)!
- n! = n(n - 1)(n - 2)................. × 1
Lets solve the problem
- Joe has one book each for algebra, geometry, history, psychology,
Spanish, English and Physics in his locker
∴ He has <em>seven</em> books in the locker
- He wants to chose <em>three</em> of them
∵ The order is not important when he chose the books
∴ We will use the combination <em>nCr</em> to find how many different sets
of three books he can choose
- The total number of books is 7
∴ n = 7
∵ He chooses 3 of them
∴ r = 3
∵ 7C3 = 7!/3!(7 - 3)! = 7!/3!(4!)
∴
∴ 7C3 = 35
* There are 35 different sets of 3 books Joe could choose
Answer:
4.5
Step-by-step explanation:
