Question

find the 40th of 5,11,17,23

Answer 1

we know that,
aN= a+(n-1)d
where,
aN=40..
a=5
d=11-5
=6
40=5+(n-1)6
40-5=(n-1)6
35/6=(n-1)
35/6+1= n
35+6/6= n
42/6=n
7=n. or n= 7

Answer 2

ok so look... this series can be treated in 2 ways

1. as an AP (arithmetic progression). this is because the difference between each number is constant-6. so by the formula Tn= a+(n-1)d if u substitute n as 40, a as 5 and d as 6... u will get Tn to be 234 which is the 40th term in the AP series.

2. as the series of alternating prime numbers, starting from 5. Here's a link to a webpage that shows primes in the first 500 numbers.

https://www.miniwebtool.com/list-of-prime-numbers/?to=500

so now, the first term in the series is the 3rd prime 5. 2nd term is the 5th prime 11. 3rd term in series is 7th prime and so on. so use the formula
prime index = 2n+1
where n is the term in that series u gave. since u want 40th term, the prime index you're looking for is 81 and the 81st prime number is 419. Then that's the answer. 419.

so 2 answers are possible by the patterns that occurred to me from the series u gave. 234 and 419.

hope u understood :) pretty cool series though.