Answer:
is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.
Step-by-step explanation:
Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.
First term of given arithmetic progression is A
and common difference is D
ie.,
and common difference=D
The nth term can be written as

pth term of given arithmetic progression is a

qth term of given arithmetic progression is b
and
rth term of given arithmetic progression is c

We have to prove that

Now to prove LHS=RHS
Now take LHS




![=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5BAq%2BpqD-Dq-Ar-prD%2BrD%5D%5Ctimes%20qr%2B%5BAr%2BrqD-Dr-Ap-pqD%2BpD%5D%5Ctimes%20pr%2B%5BAp%2BprD-Dp-Aq-qrD%2BqD%5D%5Ctimes%20pq%7D%7Bpqr%7D)




ie., 
Therefore
ie.,
Hence proved
Rate of change between 1 and 3: find the y values for 1 and 3:
Rate of change = (6-1) / (3-1) = 5/2
Rate of change from 2 and 4:
= (10-3) / (4-2) = 7/2
The rate from 2 to 4 is larger because 7/2 is larger than 5/2
The answer to this one is A
Work it exactly like the other one with Wilda and Karla.
Juan takes 15 hours to do a job ... he does 1/15 of it in an hour.
His father can do it in 6 hours ... he does 1/6 of it in an hour.
Juan works alone for 4.5 hours. In that time, he does 4.5/15 = 0.3
of the job. When his father joins him, there's only 0.7 of it to finish.
Working together, they do (1/15 + 1/6) = (2/30 + 5/30) = 7/30 of the job
each hour.
How many times do they have to do 7/30 of the job in order to finish
the 7/10 of it that remains ? That will be the number of hours they need.
(7/10) / (7/30) = 7/10 x 30/7 = <em>3 hours</em>.
Then they can knock off while it's still daylight, and go for a swim.
Answer:
a,c
Step-by-step explanation:
my brain is very large