Question

This is progression
i need to know C plsssss
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Answer 1

`l = 28`

Explanation:

Given

`S = \sum (2k - 3); k = 4\ to\ l`

Required

What is l when S = 725

This can be solved using Sum of n terms of an AP;

`S_n = (n)/(2)(T_1 + T_n)`

Where

`S_n = 725`

`T_1 = first\ term`

To get T1; we substitute 4 for k in 2k - 3

`T_1 = 2 * 4 - 3`

`T_1 = 8 - 3`

`T_1 = 5`

`T_n = last\ term`

To get Tn; we substitute l for k in 2k - 3

`T_n = 2 * l - 3`

`T_n = 2l - 3`

n = the number of terms;

Since k = 4 to l, then

`n = l - 4 +1`

`n = l - 3`

Substitute these values in
`S_n = (n)/(2)(T_1 + T_n)`

`725 = (l-3)/(2)(5 + 2l - 3)`

Collect Like Terms

`725 = (l-3)/(2)(2l + 5- 3)`

`725 = (l-3)/(2)(2l + 2)`

Open the bracket

`725 = (l-3)/(2) * 2l + (l-3)/(2) * 2`

`725 = (l-3) * l + (l-3)`

`725 = l^2-3l + l-3`

`725 = l^2-2l -3`

Subtract 725 from both sides

`725 - 725 = l^2-2l -3 - 725`

`l^2-2l -3 - 725 = 0`

`l^2-2l - 728 = 0`

`l^2 + 26l - 28l - 728 = 0`

`l(l + 26) - 28(l + 26) = 0`

`(l - 28)(l + 26) = 0`

`l - 28 = 0` or
`l + 26 = 0`

`l = 28` or
`l = -26`

But l must be positive;

Hence,
`l = 28`