The figure gives the formula for m consecutive numbers.
Here we have m+1=5 and m=4 .
In this case 265 = (5/2)x(2xn + 4) & 106 = 2xn + 4 & 2n= 102 & n=51 & the gratest of them 51+4 = 55 & 55/15 = 11 .
But this is the formal way to solve; in tests time is scarce. Look at the second tip and ry1 .
As is in ry1, in case we have an odd number of numbers, sum divided buy number gives the midddle term.
If it is demanded "k" cents for the first copy, k + 11 x (k-30) cents will be demanded for 12 copies.
Use of trial and error can laso be made:
Since this is a pretty hard nut for 4th graders, trying the easiest to handle choice we get: the first is 100 cents, and 11x70=770 cents for the others .
The sum will not be even for "b".
Numbers in "a" are not consecutive.
Let the first number be n .
The second will be 2 x (n+1) and the sum n + 2x(n+1) = 212 .
In case of trial and error:
"a" and "b" are excluded (the first tip) .
Suppose "c" is true; first no. is 80 and second term of the sum is 2x81=162. Fails.
For each natural number n: n+(n+1)+(n+2) = n+3
3xn + 3 = 5xn - 2 i.e. 2xn = 5 must ve valid but can not be for natural numbers !