Question

Q2: 8,15,22,29,36,............ A sequence of numbers is shown above. a) Find the 10th term of the sequence. *

This is a required question
b) Find the nth term of the sequence. *
c) Which term of the sequence is equal to 260. *

Answer 1

a) 71
b) an = al + (n +1) x 7
c) 37th term

Answer 2

a) 71

b) an = a1 + (n-1)×7

c) the 37th term

Explanation:

clearly, an+1 = an + 7

so, any new term is the previous term plus 7.

a1 = 8

and then the sequence goes on. starting with a2 we add 7 every time.

so, for an we added 7 n-1 times to a1. n-1 because for a1 we did not have to add 7.

therefore, the building function of this sequence to calculate the nth term is

an = a1 + (n-1)×7

a10 = a1 + 9×7 = 8 + 63 = 71

and

ax = 260 = a1 + (x-1)×7 = 8 + 7x - 7 = 1 + 7x

259 = 7x

x = 37

=> the 37th term of the sequence is 260