Question

how many terms are there in the sequence 3,6,9,12,....,111

Answer 1

15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 75 78 81 84 87 90 93 96 99 102 105 108
hope it helps

Answer 2

Let, Last term or nth term = 111



First term = 3



Common difference = any term - previous term



Here, Common difference = 6 - 3 = 9 - 6 = 12 - 9 = 3






We know,
$a_{n} = a + ( n - 1 ) d$


Applying formula,


= > 111 = 3 + ( n - 1 )3

= > 111 - 3 = ( n - 1 )3

= > 108 = ( n - 1 )3

= > 108 / 3 = n - 1

= > 36 = n - 1

= > 37 = n




Hence, 111 is the 37th term of the AP.so there are 37 terms in AP.