Question

The consecutive numbers of an arithmetic sequence are k + 1, 2k + 1, 13. Find the value of k.

Answer 1

Hi,

The consecutive numbers are 5, 9, 13

Explanation:

Let's assume a the first term,

r the common difference,

a+r the second term

a+2r the third.

`\left\{\begin{array}{ccc}a&=&k+1\\a+r&=&2k+1\\a+2r&=&13\\\end {array} \right.\\\\\\\left\{\begin{array}{ccc}a+r-a&=&2k+1-(k+1)\\a+r&=&2k+1\\a+2r&=&13\\\end {array} \right.\\\\\\\left\{\begin{array}{ccc}r&=&k\\a+k&=&2k+1\\a+2r&=&13\\\end {array} \right.\\\\\\\left\{\begin{array}{ccc}r&=&k\\a&=&k+1\\k+1+2k&=&13\\\end {array} \right.\\\\\\\left\{\begin{array}{ccc}3k+1&=&13\\a&=&k+1\\r&=&k\\\end {array} \right.\\`

`\\\\\left\{\begin{array}{ccc}k&=&4\\a&=&5\\r&=&4\\\end {array} \right.\\\\\\`

The consecutive numbers are 5, 9, 13

Answer 2

k = 4

Explanation:

there is a common difference d between consecutive terms in an arithmetic sequence , that is

d = a₂ - a₁ = a₃ - a₂ = ..... =
`a_(n)` -
`a_(n-1)` ( n is the term number )

Applying this to the 3 terms given

2k + 1 - (k + 1) = 13 - (2k + 1)

2k + 1 - k - 1 = 13 - 2k - 1 ( simplify both sides )

k = - 2k + 12 ( add 2k to both sides )

3k = 12 ( divide both sides by 3 )

k = 4

the 3 terms given are then

5, 9 , 13