Question 838068
sequence 4,12,.....972

can this be arithmetic, if so what is equation for nth term and what term number is 972<pre>If so then d = 12-4 = 8, a<sub>1</sub> = 4

a<sub>n</sub> = a<sub>1</sub>+(n-1)d

We see if 972 can be = a<sub>n</sub> for some natural number n

972 = 4 + (n-1)(8)
972 = 4 + 8n - 8
972 = 8n - 4
976 = 8n
122 = n

Yes, since 122 is a natural number, and 972 is term number 122.

So

a<sub>n</sub> = a<sub>1</sub>+(n-1)d  becomes

a<sub>n</sub> = 4+(n-1)8 = 4+8n-8 = 8n-4

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</pre>
sequence 4,12,.....972

can this be geometric, if so what is equation for nth term and what term number is 972?<pre>If so then r = 12/4 = 3, a<sub>1</sub> = 4

a<sub>n</sub> = a<sub>1</sub>r<sup>n-1</sup>

We see if 972 can be = a<sub>n</sub> for some natural number n

972 = 4·3<sup>n-1</sup>
Divide both sides by 4

243 = 3<sup>n-1</sup>

We can break 243 as 3<sup>5</sup>

3<sup>5</sup> = s<sup>n-1</sup>

So the exponents of 3 must be equal:

5 = n-1`

6 = n

Yes, since 6 is a natural number.  In fact since 6 isn't so large,
we can write the sequence to 6 terms, though it isn't necessary,
it's just a check:

4, 12, 36, 108, 324, 972

So 972 really is term number 6.

a<sub>n</sub> = a<sub>1</sub>r<sup>n-1</sup>  becomes

a<sub>n</sub> = 4·3<sup>n-1</sup>

Edwin</pre>