Question

The first term of an arithmetic sequence is 11. It's 32nd term is 600. Determine the sum of the first 50 terms

of this sequence​.

Answer 1

The sum of the first 50 terms is 23825

Step-by-step explanation:

You posted this question in the wrong subject; This is mathematics not geography; However, the solution is as follows

Given

First Term: 11

32nd Term = 600

Required

Sum of first 50 terms

Using proper notations

`a = 11\\T_(32) = 600`

First the common difference has to be calculated;

The nth term of an arithmetic sequence is as follows

`T_n = a + (n-1)d`

Where d represents the common difference

Using the data for
`T_(32)`; Substitute 11 for a and 32 for n

`T_(32) = 11 + (32-1)d`

`T_(32) = 11 + (31)d`

`T_(32) = 11 + 31d`

Recall that
`T_(32) = 600`

The above expression becomes

`600 = 11 + 31d`

Subtract 11 from both sides

`600 - 11 = 11 - 11 + 31d`

`589 = 31d`

Divide both sides by 31

`(589)/(31) = (31d)/(31)`

`(589)/(31) = d`

`19 = d`

`d = 19`

At this point the sum of first 50 terms can be calculates;

`S_n = (n)/(2)(2a + (n-1)d)`

Substitute a = 11; n = 50 and d = 19

`S_(50) = (50)/(2)(2 * 11 + (50-1)19)`

Start by solving the inner brackets

`S_(50) = (50)/(2)(2 * 11 + (49)19)`

`S_(50) = (50)/(2)(2 * 11 + 931)`

`S_(50) = (50)/(2)(22 + 931)`

`S_(50) = (50)/(2)(953)`

`S_(50) = 25(953)`

`S_(50) = 23825`

Hence, the sum of the first 50 terms is 23825