Arithmetic Progression: Finding the Value of k

The problem involves an arithmetic progression (AP) where the sum of terms up to the last term, represented by '$k$', is given as -105. We need to find the value of '$k$'.

Identifying AP Parameters

• The first term ($a$) is 15.
• The common difference ($d$) is $13 - 15 = -2$.
• The sum of the series ($S_n$) is -105.
• The last term ($a_n$) is $k$.

Calculating the Number of Terms (n)

We use the formula for the sum of an AP:

$ S_n = \frac{n}{2}[2a + (n-1)d] $

Substitute the known values:

$ -105 = \frac{n}{2}[2(15) + (n-1)(-2)] $

Simplify the equation:

$ -105 = \frac{n}{2}[30 - 2n + 2] $

$ -105 = \frac{n}{2}[32 - 2n] $

$ -105 = n(16 - n) $

$ -105 = 16n - n^2 $

Rearrange into a quadratic equation:

$ n^2 - 16n - 105 = 0 $

Factor the quadratic equation:

$ (n - 21)(n + 5) = 0 $

The possible values for '$n$' are $n = 21$ or $n = -5$. Since the number of terms cannot be negative, we take $n = 21$.

Determining the Last Term (k)

Now, we find the value of the last term '$k$', which is the 21st term ($a_{21}$), using the formula for the nth term of an AP:

$ a_n = a + (n-1)d $

Substitute $n = 21$, $a = 15$, and $d = -2$:

$ k = a_{21} = 15 + (21-1)(-2) $

$ k = 15 + (20)(-2) $

$ k = 15 - 40 $

$ k = -25 $

Therefore, the value of '$k$' is -25.