Given that the common difference is 3, find the 56th term and also the sum of its 32terms
To find the 56th term of the arithmetic sequence with a common difference of 3, we can use the formula for the nth term of an arithmetic sequence:
a_n = a_1 + (n - 1)d
Where:
a_n is the nth term
a_1 is the first term
d is the common difference
n is the term number
Given that the common difference is 3 and a_1 = 1, we can find the 56th term:
a_56 = 1 + (56 - 1) * 3
a_56 = 1 + 165
a_56 = 166
Therefore, the 56th term of the arithmetic sequence is 166.
To find the sum of the first 32 terms of the arithmetic sequence, we can use the formula for the sum of an arithmetic series:
S = n/2 * (2a_1 + (n - 1)d)
Where:
S is the sum of the series
n is the number of terms
a_1 is the first term
d is the common difference
Given that the common difference is 3, the first term is 1, and the number of terms is 32, we can find the sum:
S = 32/2 * (2*1 + (32 - 1) * 3)
S = 16 * (2 + 93)
S = 16 * 95
S = 1520
Therefore, the sum of the first 32 terms of the arithmetic sequence is 1520.