Find the number of terms in the sequence $4, 12, 20.......108$.
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Find the number of terms in the sequence $4, 12, 20.......108$.
This is an arithmetic progression with a = 4, d = 8, and last term l = 108. Formula: l = a + (n-1)d. 108 = 4 + (n-1)8. 104 = (n-1)8. 13 = n-1, so n = 14.
The sequence is an arithmetic progression with first term 4 and common difference 8. The nth term formula is a + (n - 1)d. Setting the last term to 108 gives 4 + (n - 1)*8 = 108, so 8(n - 1) = 104 and n - 1 = 13. This means n equals 14, so there are 14 terms in the sequence.