n(n+1)
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The difference between the terms increases by 2 each time. 20-12=8 30-20=10 42-30=12 56-42=14 ... f(n)=(n+2)(n+3)
here t1=2 t2=6 t3=12 t4=20 t5=30 the nth term is n(n+1) for t1 = 1(1+1)=2 for t2=2(2+1)=6 for t3=3(3+1)=12 for t4=4(4+1)=20 for t5=5(5+1)=30 for tn=n(n+1)
Looks like 57: 12+9=21, +9=30, +9=39, +9=48, +9=57.
6n+10
Well, honey, if the nth term is 3n-1, then all you gotta do is plug in n=30 and do the math. So, the 30th term would be 3(30)-1, which equals 89. There you have it, sweet cheeks, the 30th term of that sequence is 89.