Answer:
114
Step-by-step explanation:
The question relates to the sum of an arithmetic progression, A. P.
The count of numbers added, n = 50
The sum of the numbers, Sₙ = 3,250
The formula for the sum of an A. P. is given as follows;
[tex]S_n = \dfrac{n}{2} \left[2 \cdot a + (n - 1)\cdot d\right][/tex]
[tex]S_n = \dfrac{n}{2} \left[a + a_n\right][/tex]
Where;
a = The first term of set of numbers
d = The common difference between consecutive numbers
n = The number of terms = 50
aₙ = The last term, the largest of the integers
Therefore, we get;
The common difference of consecutive even integers numbers, d = 2
Plugging in the values gives;
3,250 = (50/2) × (2·a + (50 - 1) × 2) = 25 × (2·a + 49 × 2)
2·a = (3,250/25) - 49 × 2 = 32
a = 32/2 = 16
From [tex]S_n = \dfrac{n}{2} \left[a + a_n\right][/tex], we have;
3,250/25 = 16 + aₙ
aₙ = 3,250/25 - 16 = 114
The largest of the integers, aₙ = 114.
