Generate an image showing a visual representation of an arithmetic progression on a number line. Show some distinct points marked along the line indicating the 5th term as 18 and the 10th term as -2. However, ensure to not include any explicit numerical values or text in the image. Visually depict the decreasing value of the terms, suggesting the progression's negative common difference.

The 5th and 10th terms of an arithmetic progression are 18 and -2 respectively.

(a) Find the common difference and the first term.
(b) Determine the least number of terms which must
be added together so that the sum of the progression is negative .Hence find the sum.

Using our definitions:

a + 4d = 18
a + 9d = -2
subtract:
5d = -20
d = -4
mentally in a + 4d = 18, a = 34 <----- first term

so we want S(n) < 0, let's see when it is zero
(n/2)(2a + (n-1)d) = 0
n/2)(68 - 4(n-1)) = 0
n/2(68 - 4n + 4) = 0
n(36 - 2n) = 0
-2n^2 + 36n = 0
n^2 - 18n = 0
n = 0 or n = 18

So when we add the first 18 terms we get zero, the terms are getting smaller
so when we add 19 terms we get a negative sum

(a) To find the common difference (d), we can use the formula:

Tn = a + (n - 1)d

where Tn is the nth term, a is the first term, n is the position of the term, and d is the common difference.

Using the information given:
T5 = 18
T10 = -2

Plug in the values into the formula for Tn and set up two equations:

18 = a + 4d ...(1)
-2 = a + 9d ...(2)

Now, we can solve these equations simultaneously to find the values of a and d.

Multiplying equation (1) by 9 and equation (2) by 4 to eliminate a:

162 = 9a + 36d ...(3)
-8 = 4a + 36d ...(4)

Subtract equation (4) from equation (3):

170 = 5a
a = 34

Now substitute the value of a back into either equation (1) or (2):

18 = 34 + 4d
4d = -16
d = -4

The common difference (d) is -4 and the first term (a) is 34.

(b) To determine the least number of terms that must be added together so that the sum of the progression is negative, we need to find the sum of the progression.

The sum of an arithmetic progression is given by the formula:

Sn = (n/2)(2a + (n-1)d)

Given that the sum should be negative, we set up the equation:

Sn < 0

Substituting the values of a and d:

(n/2)(2(34) + (n-1)(-4)) < 0

Simplifying the equation:
(n/2)(68 - 4n + 4) < 0
(n/2)(72 - 4n) < 0
(n(72 - 4n))/2 < 0
n(72 - 4n) < 0

Now we can find the values of n that satisfy this inequality. Let's plot it on a number line:

0-----4------18-----18-----36

Looking at the intervals, we can see that n must be between 0 and 9 for the sum to be negative.

Therefore, the least number of terms that must be added together is 9. To find the sum, plug in n = 9 into the formula for Sn:

S9 = (9/2)(2(34) + (9-1)(-4))
S9 = 9(68 - 36)
S9 = 9(32)
S9 = 288

So, the sum of the arithmetic progression is 288.

To find the common difference and the first term of the arithmetic progression, we can use the formula for the nth term of an arithmetic progression:

\(a_n = a_1 + (n-1)d\)

where \(a_n\) represents the nth term, \(a_1\) represents the first term, \(n\) represents the term number, and \(d\) represents the common difference.

(a) Given that the 5th term is 18 and the 10th term is -2, we can plug these values into the formula:

For the 5th term:
\(a_5 = a_1 + (5-1)d = 18\)

For the 10th term:
\(a_{10} = a_1 + (10-1)d = -2\)

We now have a system of equations with two unknowns (the common difference, \(d\), and the first term, \(a_1\)):

\(a_1 + 4d = 18\)
\(a_1 + 9d = -2\)

To solve this system, we can subtract the first equation from the second equation to eliminate \(a_1\):

\(a_1 + 9d - (a_1 + 4d) = -2 - 18\)
\(5d = -20\)

Dividing both sides by 5 gives us \(d = -4\).

Now we can substitute this value of \(d\) into one of the original equations to find the first term. Let's use the equation for the 5th term:

\(a_1 + 4(-4) = 18\)
\(a_1 - 16 = 18\)
\(a_1 = 34\)

Therefore, the common difference is -4 and the first term is 34.

(b) To find the least number of terms that must be added together so that the sum of the progression is negative, we can use the formula for the sum of the first \(n\) terms of an arithmetic progression:

\(S_n = \frac{n}{2}(2a_1 + (n-1)d)\)

We want to find the smallest value of \(n\) such that the sum \(S_n\) is negative. Let's set up the equation:

\(\frac{n}{2}(2(34) + (n-1)(-4)) < 0\)

\(\frac{n}{2}(68 - 4n + 4) < 0\)

\(n(72 - 4n) < 0\)

Simplifying further:

\(72n - 4n^2 < 0\)

Rearranging the terms:

\(4n^2 - 72n > 0\)

Factoring out a common factor of 4:

\(4n(n - 18) > 0\)

To find the values of \(n\) that satisfy this inequality, we can set each factor equal to zero:

\(4n = 0\) or \(n - 18 = 0\)

This gives two potential solutions: \(n = 0\) and \(n = 18\).

However, the sum of 0 terms is 0, and the sum of 18 terms is positive. Therefore, neither of these values satisfies our condition of finding the smallest value of \(n\) such that the sum is negative.

In this case, there is no smallest positive integer \(n\) that will make the sum negative. The arithmetic progression will always have a positive sum.

Therefore, the sum of the progression is always positive.

To solve this problem, we can use the formula for arithmetic progression where the nth term of the AP is given by:

An = A1 + (n-1)d

where An is the nth term, A1 is the first term, n is the term number, and d is the common difference.

(a) Find the common difference and the first term:

Given: A5 = 18, A10 = -2

To find the common difference, we will subtract A5 from A10:

-2 - 18 = -20

Therefore, the common difference (d) is -20.

To find the first term, we can substitute the values of A5 and d into the formula:

A5 = A1 + (5-1)d
18 = A1 + 4(-20)
18 = A1 - 80
A1 = 18 + 80
A1 = 98

Therefore, the first term (A1) is 98.

(b) Determine the least number of terms which must be added together so that the sum of the progression is negative. Hence, find the sum:

We need to find the number of terms (n) such that the sum of the arithmetic progression is negative. The sum of an arithmetic progression is given by:

Sn = (n/2)(2A1 + (n-1)d)

We want to find the smallest value of n such that Sn is negative. Let's substitute the values of A1, d, and Sn into the formula and solve for n:

Sn = (n/2)(2(98) + (n-1)(-20))

Since we don't know the value of n yet, we'll assume that the sum is negative and solve for n.

Let's say Sn is negative, so Sn < 0:

(n/2)(2(98) + (n-1)(-20)) < 0

Multiplying through by 2 will give:

n(196 - 20n + 20) < 0

Simplifying:

n(216 - 20n) < 0

We have a quadratic inequality, and we can solve this by finding the critical points:

n(216 - 20n) = 0

Setting each factor equal to zero and solving:

n = 0 or 216 - 20n = 0

The first solution, n = 0, is not valid because it is not a positive integer.

Solving the second equation:

216 - 20n = 0
20n = 216
n = 216/20
n = 10.8

The number of terms, n, cannot be a decimal, so we take the next higher integer, which is 11.

Therefore, the least number of terms that must be added together so that the sum of the progression is negative is 11.

To find the sum, we can substitute the values of A1, d, and n into the formula for the sum:

Sn = (n/2)(2A1 + (n-1)d)
S11 = (11/2)(2(98) + (11-1)(-20))
S11 = (11/2)(196 - 200)
S11 = (11/2)(-4)
S11 = -22

Therefore, the sum of the arithmetic progression is -22.