Find the 25th term of this arithmetic sequence: 132, 136, 140, 144,...

Answer 1

Subtract the first term from the second term and the second term from the third term. Each answer is 4. That tells you that it is an arithmetic sequence with a common difference of 4.

Then use the formula a sub n = a sub 1 + d( n- 1)

a sub 25 = 132 + 4 ( 25-1 )

a sub 25 = 132 + 4 ( 24 )

a sub 25 = 132 + 96

Then a sub 25 = 228

Answer 2

a1 = 132

a2 = 136

a3 = 140

a4 = 144

a5 = 148

a6 = 152

every 5 terms the term ends in 2

terms end in a a last digit pattern, 2,6,0,4, 8, then repeat,

eat, sleep, wake up, repeat,

shampoo, wash, rince, repeat

common difference = d = 4

nth term = an = a1 + d(n-1) = 132 +4(n-1)

a25 = 132 +4(25-1) = 132 +4(24) = 132 + 96 = 228

25th term is 228

25/5 = 5 cycles of 5 terms, endng in 8

2,6,0,4,8, repeat

21st, 22nd, 23rd, 24th, 25th terms are

212, 216, 220, 224, 228, ...