Quadratic Sequence and Geometric Sequence

In a geometric sequence, divide a term by the one before it to find the common ratio $r$.

$10÷5=2$, and checking again, $20÷10=2$.

The ratio is the same each time, so the common ratio is 2.

First find the common ratio by dividing a term by the one before it: $32÷64=0.5$.

The ratio is less than 1, so the terms get smaller each time.

To find the next term, multiply the last term by the ratio: $8\times 0.5=4$.

So the next term is 4.

A quadratic sequence has a constant second difference, so first find the first differences.

$4-3=1$, $7-4=3$, $12-7=5$, $19-12=7$, giving $1,3,5,7$.

Now take the differences of those: $3-1=2$, $5-3=2$, $7-5=2$.

So the second difference is constant at 2.

Common mistake: the answer is the difference of the first differences, not a first difference itself.

First find the first differences: $8-5=3$, $14-8=6$, $23-14=9$, $35-23=12$.

The second difference is constant: $6-3=3$, so it stays at 3.

Add the second difference to the last first difference: $12+3=15$.

Then add this to the last term: $35+15=50$.

So the next term is 50.

The nth term of a geometric sequence is $a\times r^{n-1}$, where $a$ is the first term and $r$ is the common ratio.

Here $a=3$, $r=2$ and $n=6$, so the 6th term is $3\times 2^{6-1}=3\times 2^5$.

$2^5=32$, so $3\times 32=96$.

So the 6th term is 96.

Common mistake: the exponent is $n-1$, not $n$, so use $2^5$ rather than $2^6$.

Find the first differences: $7-4=3$, $14-7=7$, $25-14=11$, $40-25=15$.

The second difference is constant at 4.

For the 6th term, the next first difference is $15+4=19$, so the 6th term is $40+19=59$.

For the 7th term, the next first difference is $19+4=23$, so the 7th term is $59+23=82$.

So the 7th term is 82.