If the sum of four consecutive numbers is 854, what is the third one?

Mathematics

1 answer:

lukranit [14]3 years ago

7 0

Answer:

214

Step-by-step explanation:

The set up would be x + (x + 1) + (x + 2) + (x + 3) = 854, naturally as the numbers are consecutive. Solving for x:

x + (x + 1) + (x + 2) + (x + 3) = 854,

x + x + 1 + x + 2 + x + 3 = 854,

4x + 1 + 2 + 3 = 854,

4x + 6 = 854,

4x = 854 - 6 = 848,

x = 848/4 = 212

The third number then should be 212 + 2 = 214

You might be interested in

I hate this!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Klio2033 [76]

Answer:

Same!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

5 0

3 years ago

Determine the total number of proper subsets of the set of letters from the English alphabet a,b, c, ...,j. The number of proper

Inga [223]

$\{a,b,c,\ldots,j\}$ has ten elements, so any proper subset has at most nine of these elements.

The number of ways of taking any $0\le n\le 9$ letters from this set is given by the binomial coefficient,

$\dbinom{10}n=\dfrac{10!}{n!(10-n)!}$

and in particular, the total number of ways to picking proper subsets is

$\displaystyle\sum_{n=0}^9\binom{10}n=\dbinom{10}0+\dbinom{10}1+\cdots+\dbinom{10}9$

Without computing each term directly, let's instead use a direct result from the binomial theorem, which says

$\displaystyle\sum_{n=0}^N\binom Nna^nb^{N-n}=(a+b)^N$

If we replace $a=b=1$ , then we're left with

$\displaystyle\sum_{n=0}^N\binom Nn=(1+1)^N=2^N$

We can use this to evaluate our sum directly:

$\displaystyle\sum_{n=0}^9\binom{10}n=\sum_{n=0}^{10}\binom{10}n-\binom{10}{10}=2^{10}-1=1023$