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

Answer:

The sum of the first 50 terms is 23825

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

$\frac{589}{31} = \frac{31d}{31}$

$\frac{589}{31} = d$

$19 = d$

$d = 19$

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

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

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

$S_{50} = \frac{50}{2}(2 * 11 + (50-1)19)$

Start by solving the inner brackets

$S_{50} = \frac{50}{2}(2 * 11 + (49)19)$

$S_{50} = \frac{50}{2}(2 * 11 + 931)$

$S_{50} = \frac{50}{2}(22 + 931)$

$S_{50} = \frac{50}{2}(953)$

$S_{50} = 25(953)$

$S_{50} = 23825$

Hence, the sum of the first 50 terms is 23825