Answer:
s + r - l
Step-by-step explanation:
This is an arithmetic progression question.
The formula to get the terms of an AP is;
L_n = a + (n - 1)d
Where;
a is first term
n is number of term
d is difference between terms
Thus, second term is;
L_2 = a + d
Third term is;
L_3 = a + (3 - 1)d
L_3 = a + 2d
Similarly, L_4 = a + 3d
Now we are told that;
Sum of the first 2 numbers is s;
Sum of the second and third number is l
Sum of the last two numbers is r
Thus;
a + (a + d) = s - - - (eq 1)
(a + d) + (a + 2d) = l - - - (eq 2)
(a + 2d) + (a + 3d) = r - - - (eq 3)
Thus, sum of the first and last term will be;
L_1 + L_4 = a + (a + 3d)
Now,from earlier;
(a + 2d) = r - (a + 3d)
Also;
(a + d) = s - a
Thus, plugging them into the eq 2 gives;
(s - a) + r - (a + 3d) = l
Expanding we have;
s + r - 2a - 3d = l
d = (s + r - 2a - l)/3
Thus;
L_1 + L_4 = a + (a + (3(s + r - 2a - l)/3))
= 2a + s + r - 2a - l = s + r - l