What is the 55th term of a sequence if the 6th term is 4 and the 100th term is 98?
What is the 55th term of a sequence if the 6th term is 4 and the 100th term is 98 ?
an=a1⋅rn−1a6=a1⋅r5=4a100=a1⋅r99=98a100a6=a1⋅r99a1⋅r5=984a1⋅r99a1⋅r5=984r99r5=24.5r99−5=24.5r94=24.5r=94√24.5r=1.03461402806a6=a1⋅r5=4a1⋅r5=4a1=4r5a1=41.034614028065a1=3.37417946747a55=a1⋅r54a55=3.37417946747⋅1.0346140280654a55=21.1933572378
If this is an arithmetic sequence then the n'th term is a + (n-1)d where a is the first and d is the difference.
a + 5d = 4
a + 99d = 98
so a = -1 and d = 1
the 55th term is -1 + 54*1 = 53
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If this is a geometric sequence then the n'th term is a*rn-1 where r is the multiplier.
a*r5 = 4
a*r99 = 98
r=(98/4)1/94 ≈ 1.0346
a = 4/1.03465 ≈ 3.3742
The 55th term is 3.3742*1.034654 ≈ 29.193
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What is the 55th term of a sequence if the 6th term is 4 and the 100th term is 98 ?
an=a1⋅rn−1a6=a1⋅r5=4a100=a1⋅r99=98a100a6=a1⋅r99a1⋅r5=984a1⋅r99a1⋅r5=984r99r5=24.5r99−5=24.5r94=24.5r=94√24.5r=1.03461402806a6=a1⋅r5=4a1⋅r5=4a1=4r5a1=41.034614028065a1=3.37417946747a55=a1⋅r54a55=3.37417946747⋅1.0346140280654a55=21.1933572378