Answer:
[tex]l = 28[/tex]
Step-by-step explanation:
Given
[tex]S = \sum (2k - 3); k = 4\ to\ l[/tex]
Required
What is l when S = 725
This can be solved using Sum of n terms of an AP;
[tex]S_n = \frac{n}{2}(T_1 + T_n)[/tex]
Where
[tex]S_n = 725[/tex]
[tex]T_1 = first\ term[/tex]
To get T1; we substitute 4 for k in 2k - 3
[tex]T_1 = 2 * 4 - 3[/tex]
[tex]T_1 = 8 - 3[/tex]
[tex]T_1 = 5[/tex]
[tex]T_n = last\ term[/tex]
To get Tn; we substitute l for k in 2k - 3
[tex]T_n = 2 * l - 3[/tex]
[tex]T_n = 2l - 3[/tex]
n = the number of terms;
Since k = 4 to l, then
[tex]n = l - 4 +1[/tex]
[tex]n = l - 3[/tex]
Substitute these values in [tex]S_n = \frac{n}{2}(T_1 + T_n)[/tex]
[tex]725 = \frac{l-3}{2}(5 + 2l - 3)[/tex]
Collect Like Terms
[tex]725 = \frac{l-3}{2}(2l + 5- 3)[/tex]
[tex]725 = \frac{l-3}{2}(2l + 2)[/tex]
Open the bracket
[tex]725 = \frac{l-3}{2} * 2l + \frac{l-3}{2} * 2[/tex]
[tex]725 = (l-3) * l + (l-3)[/tex]
[tex]725 = l^2-3l + l-3[/tex]
[tex]725 = l^2-2l -3[/tex]
Subtract 725 from both sides
[tex]725 - 725 = l^2-2l -3 - 725[/tex]
[tex]l^2-2l -3 - 725 = 0[/tex]
[tex]l^2-2l - 728 = 0[/tex]
[tex]l^2 + 26l - 28l - 728 = 0[/tex]
[tex]l(l + 26) - 28(l + 26) = 0[/tex]
[tex](l - 28)(l + 26) = 0[/tex]
[tex]l - 28 = 0[/tex] or [tex]l + 26 = 0[/tex]
[tex]l = 28[/tex] or [tex]l = -26[/tex]
But l must be positive;
Hence, [tex]l = 28[/tex]
