What is the 32nd term of the arithmetic sequence where a1 = 12 and a13 = –60 pleaseee help me figure this out!
arithmetic sequence is one where each term is obtained by previous by adding a constant number called the common difference so if a is the first term the sequence is a, a+2, a+2d, a+3d ........ where a is common difference
a13 the 13th term = a + 12d = -60 a1 first term = 12 so ( a + 12d) - a = -60 -12 = -72 so 12d = -72 d = -6 32nd term = a + 31d = 12 + (-6*31)
What is the 32nd term of the arithmetic sequence where a1 = –33 and a9 = –121 whats d in this problem
go on - try this one on your own! just go through the same process as i did in the last post a1 = -33 a9 = a1 + 8d = -121 and so on
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