Question 1151584


Find the {{{100}}}th and nth term for each of the following sequences.
a. 1,5,9,13,...

{{{a[1]=1}}}
{{{a[2]=5=1+4}}}
{{{a[3]=9=5+4}}}
{{{a[4]=13=9+4}}}

=> common difference {{{d=4}}}, arithmetic sequence
 
the {{{n}}}th formula:

{{{a[n]= a[1]+(n-1)d}}}

so,  the {{{100}}}th term will be:

{{{a[100]= 1+(100-1)4}}}

{{{a[100]= 1+99*4}}}

{{{a[100]= 1+396}}}

{{{a[100]= 397}}}



b.70,120,170...

{{{a[1]=70}}}
{{{a[2]=120=70+50}}}
{{{a[3]=170=120+50}}}

=> common difference {{{d=50}}}, arithmetic sequence
 
the {{{n}}}th formula:

{{{a[n]= a[1]+(n-1)d}}}

so,  the {{{100}}}th term will be:

{{{a[100]= 70+(100-1)50}}}

{{{a[100]= 70+99*50}}}

{{{a[100]= 70+4950}}}

{{{a[100]= 5020}}}


c.1,3,9....

{{{a[1]=1}}}
{{{a[2]=3=1*3}}}
{{{a[3]=9=1*3^2}}}

=> common ratio {{{r=3}}}, geometric sequence

the {{{n}}}th formula:

 {{{a[n] = a[1]*3^(n - 1)}}}

the {{{100}}}th term will be:

{{{a[100] = 1*3^(100 - 1)}}}
{{{a[100] = 1*3^99}}}
{{{a[100] = 171792506910670443678820376588540424234035840667}}}


d.

8,8^4,8^7,8^10,...

note that exponents of first and second term differ by {{{3}}}, same for second and third term, and same for third and fourth term

 {{{a[1]=8^1}}} 
 {{{a[2]=8^4=8^(1+3)}}} 
 {{{a[2]=8^7=8^(4+3)}}}
 
the {{{n}}}th formula:

 {{{a[n] =8^(3n - 2)}}} where {{{n}}}={{{0}}},{{{1}}},....

the {{{100}}}th term will be:

{{{a[100] =8^(3*100 - 2)}}}
{{{a[100] =8^(3*98)}}}
{{{a[100] =8^294}}} => leave like this because this number is really big


e.

109+7x3^32,109+8x3^32,109+9x3^32,... 

I am not quite sure  if x represents, I assume multiplication


{{{109+7*3^32}}},{{{109+8*3^32}}},{{{109+9*3^32}}}


only difference is in numbers {{{7}}},{{{8}}},{{{9}}}


generally, {{{(n+7)}}} where {{{n}}}= {{{0}}},{{{1}}},....
 
the {{{n}}}th formula:

 {{{a[n] =109+(n+7)*3^32}}} 

 {{{a[100] =109+(100+7)*3^32}}} 

 {{{a[100] =109+107*3^32 }}}

{{{a[100] =198273160207147096}}}