Suppose n=6 , m=2 , z1 is the average of the elements of x , and z2 is the average of the first three elements of x minus the average of fourth through sixth elements of x . Determine A . Note: Enter A in a list format: [[A11,...,A16],[A21,...,A26]]

See step by step explanations fir answers.

Step-by-step explanation:

Given that;

n=6 , m=2 , z1 is the average of the elements of x , and z2 is the average of the first three elements of x minus the average of fourth through sixth elements of x

So, Let d=n∧28, d′=n∧22.

We have nd=10, nd′=20n

so

n=10d=20d′nwhenced=2d′.

On the other hand, d is a divisor of 28, and above shows d′ is, too. As it is also a divisor of 22, the only possibilities are d′=1, d′=2, corresponding to d=2, d=4.

However, if d=2, n=20, and 20∧28=4, not 2. So the only solution is d=4, and n=40.

= 5

step-by-step explanation:

to find the 15th term substitute n = 15 into the nth term formula

= (- 3 × 15 ) + 50 = - 45 + 50 = 5