The fourth term of the arithmetic progression is 16, and the sum of the sixth and eleventh is 5.

The fourth term of the arithmetic progression is 16, and the sum of the sixth and eleventh is 5. Find the sum of the first 18 terms of the progression.

find the first term and the difference of the given arithmetic progression.

Let’s use the formula of the nth term of the arithmetic progression an = a1 + (n – 1) * d, where a1 is the first term of the arithmetic progression, d is the difference of the arithmetic progression.

By the condition of the problem, the fourth term a4 of this arithmetic sequence is 16, and the sum of the sixth and eleventh terms is 5, therefore, we can write the following ratios:

a1 + (4 – 1) * d = 16;

a1 + (6 – 1) * d + a1 + (11 – 1) * d = 5.

Simplifying these ratios, we get:

a1 + 3d = 16;

2a1 + 15d = 5.

We solve the resulting system of equations.

Substituting into the second equation the value a1 = 16 – 3d from the first equation, we get:

2 * (16 – 3d) + 15d = 5;

32 – 6d + 15d = 5;

32 + 9d = 5;

9d = 5 – 32;

9d = -27;

d = -27 / 3;

d = -9.

Find a1:

a1 = 16 – 3d = 16 – 3 * (-9) = 16 + 27 = 43.

Find the sum of the first 18 members of the progression:

S16 = (2 * a1 + d * (16 – 1)) * 16/2 = (2 * a1 + d * 15) * 8 = (2 * 43 + (-9) * 15) * 8 = (86 – 135) * 8 = -49 * 8 = -392.

Answer: The sum of the first 18 members of the progression is 392.