Question

Find the value of x if 1 + 6 + 11 + ... + x = 189.

Answer 1

1 + 6 + 11 + ... + x = 189

a = 1

d = 6 − 1

= 5

a_n = x

S_n = 189

`n/2[2a + (n - 1)d] = 189`

`n/2[2 + (n - 1)5] = 189`

`n/2[2 + 5n - 5] = 189`

`n/2[5n - 3] = 189`

5n² − 3n = 189 × 2

5n² − 3n = 378

5n² − 3n − 378 = 0

n = `(−(−3) ± sqrt(9 - 4(5)(-378)))/(2(5))`

= `(3 ± sqrt(9 + 7560))/10`

= `(3 ± sqrt(7569))/10`

= `(3 ± 87)/10`

= `(3 + 87)/10, (3 - 87)/10`

= `90/10, (-84)/10`

n = 9

a_n = a + (n −1 )d

x = 1 + (9 − 1)5

x = 1 + 8 × 5

x = 1 + 40

x = 41