Question

6,10,14 .... is an arithmetic sequence. Where is the position of the term 402 in the sequence?

Answer 1

Answer:

Given

a = 6

d = 10-6 = 4

Tn= a+ (n-1)d

402 = 6+(n-1)4

402 = 6+4n-4

400= 4n

n= 400/4

n = 100

So the answer is 109

Answer 2

$a = 6 \: \: and \: \: d = 4$

$aₙ = 402$

$a + (n - 1)d = 402$

$6 + (n - 1)4 = 402$

$(n - 1)4 = 402 - 6$

$n - 1 = \frac{396}{4}$

$n = 99 + 1$

$n = 100$