Find the 35th term of the given sequence. 3, 7, 11, 15, …
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Find the 35th term of the given sequence. 3, 7, 11, 15, …
130
133
139
141
This is an arithmetic progression with a=3 and d=4. The n-th term is a + (n-1)d. 35th term = 3 + (34 * 4) = 3 + 136 = 139.
The given sequence is an arithmetic progression with the first term a as 3 and the common difference d as 4. Using the formula for the nth term of an arithmetic progression, Tn = a + (n - 1)d, we substitute the known values to find the 35th term: T35 = 3 + (35 - 1)4. This simplifies to 3 + 136, resulting in 139.