Question

Find the 100th AND the nth term for the following sequence. Please show work.

Answer 1

Answer:

nth term of this sequence is $(197+(n+6)\times 3^{27})$

and 100th term is $(197+106\times 3^{27})$.

Step-by-step explanation:

The given sequence is $(197+7\times 3^{27}),(197+8\times 3^{27}),(197+9\times 3^{27})$

Now we will find the difference between each successive term.

Second term - First term = $(197+8\times 3^{27})-(197+7\times 3^{27})$

                                         = $(8\times 3^{27}-7\times 3^{27})$

                                         = $3^{27}(8-7)$

                                         = $3^{27}$

Similarly third term - second term = $3^{27}$

So there is a common difference of $3^{27}$.

It is an arithmetic sequence for which the explicit formula will be

$T_{n}$=a + (n - 1)d

where $T_{n}$ = nth term of the arithmetic sequence

where a = first term of the arithmetic sequence

n = number of term

d = common difference in each successive term

Now we plug in the values to get the 100th term of the sequence.

$T_{n}=(197+7\times 3^{27})+(n-1)\times 3^{27}$

               = $(197+(n+6)\times 3^{27})$

$T_{100}=(197+7\times 3^{27})+(100-1)\times 3^{27}$

                   = $197+7\times 3^{27}+99\times 3^{27}$

                   = $197+106\times 3^{27}$

Therefore, nth term of this sequence is $(197+(n+6)\times 3^{27})$

and 100th term is $(197+106\times 3^{27})$.