Minimum Arithmetic Progression?

subsitute a for the separation between subsequent variables. $x, x+a, x+2*a, x+3*a, x+4*a, x+5*a$ this sum must adds to $6*x + 15*a=78$ . You have to pick an integer x such that 15 divides $(78-6*x)$ evenly. $(78-6*x)=0\ \ mod \ 15$ . Substituting x=1 and x=2 doesn't allow for integer solutions, x=3 => a=4 Q.E.D. since any greater spacing will lead to a sum > 78

Let the first term of the arithmetic progression be 'a' . Let the difference between the terms be 'd'.

We have the formula for the sum of n terms of an AP :

Sn = $\frac {n}{2}$ [2a +(n-1)d]

Here , n = 6 , Sn = 78

Substituting , we get :

78 = $\frac{6}{2}$ [2a + 5d}

78 = 3[2a + 5d]

26 = 2a + 5d

As a and d are positive integers , the minimum value a can take is 3

Therefore , 5d = 20

d = 4

a1 + aN = 26 via n/2 * (a1+aN)d
x+y = 26 eq ..1
x-y = 10 or 20
Therefore d (difference ) can be 2 or 4

$a+(a+b)+...+(a+5b) = 6a+30b = 78 \leftrightarrow 2a+5b = 26$

this is a diophantine equation, which for maximum of b reach when a is lowest, but we want a and b to be integer. We will find that the maximum of b is reached when $a = 3 \rightarrow b = 4$

Average of six terms = $\frac{78}{6}$ = 13.

Average is also equal to smallest term (let it be a) + 2.5d where d is the common difference.

So a + 2.5d = 13

For d to be an integer, smallest a = 3

Hence 2.5d =10 and d=4 yay.

Given: the sum of an arithmetic progression of six positive integers is 78.

Find: the greatest possible difference between consecutive terms.

Let x = the first term.

Let y = the difference between terms.

Then the terms are x, x + y, x + 2y, x + 3y, x + 4y, x + 5y

x + x + y + x + 2y + x + 3y + x + 4y + x + 5y = 6x + 15y = 78

2x + 5y = 26

We are looking for the largest value of y. Let’s start with x = 1.

2+5y = 26

5y = 24

y would not be an integer. For y to be an integer, the value on the right-hand side of the equation must be divisible by 5. Instead of 2 (for 2x) we need 6 to be the value of 2x, or x = 3.

6+5y = 26

5y = 20

y = 4 --> Ans.