Question

How many terms must be added in an arithmetic sequence whose first term is 1818 and whose common difference is 44 to obtain a sum of 40184018​?

Answer 1

$T_n = a_1 + d(n - 1)$

When a = 18, d = 4, find the nth sequence:

$T_n = 18 + 4(n - 1)$

$T_n = 18 + 4n - 4$

$T_n = 14 + 4n$

When the nth term is 4018, find the value of n:

$14 + 4n = 4018$

$4n = 4018 - 14$

$4n = 4004$

$n = 1001$

Answer: 1001 term must be added to obtain the sum of 4018.