Question 1160078


1. Find the {{{25}}}th term of the arithmetic sequence
{{{-7}}}, {{{-4}}}, {{{-1}}}, {{{2}}}, ...

first term: {{{a[1]= -7}}}
second term: {{{a[2]=-7+3=-4}}}
third term:{{{a[3]=-4+3=-1}}}
fourth term: {{{a[4]=-1+3=2}}}

so, common difference is {{{d=3}}}

nth term formula is:

{{{a[n]=a[1]+d(n-1)}}}
{{{a[n]=-7+3(n-1)}}}


{{{25}}}th term->{{{n=25}}}

{{{a[25]=-7+3(25-1)}}}
{{{a[25]=-7+3(24)}}}
{{{a[25]=-7+72}}}
{{{a[25]=65}}}


2. In an arithmetic sequence, if {{{a[4] = 18}}} and {{{a[10] = 30}}}, determine {{{a[1]}}}, {{{d}}}, and {{{a[n]}}}.

Then write the first four terms of the sequence.

{{{a[n]=a[1]+d(n-1)}}}....use {{{a[4] = 18}}}

{{{18=a[1]+d(4-1)}}}
{{{18=a[1]+3d}}}...........solve for {{{a[1]}}}
{{{18-3d=a[1]}}}..........eq.1

{{{a[n]=a[1]+d(n-1)}}}....use {{{a[10] = 30}}}
{{{30=a[1]+d(10-1)}}}
{{{30=a[1]+9d}}}
{{{30-9d=a[1]}}}..............eq.2

from eq.1 and eq.2 we have

{{{18-3d=30-9d}}}......solve for {{{d}}}

{{{9d-3d=30-18}}}
{{{6d=12}}}
{{{d=12/6}}}
{{{d=2}}}

go to eq.1
{{{18-3d=a[1]}}}..........eq.1, substitute {{{d}}}
{{{18-3(2)=a[1]}}}
{{{18-6=a[1]}}}
{{{12=a[1]}}}

your nth term formula is:

{{{a[n]=12+2(n-1)}}}

since first term is {{{a[1]=12}}} and  the difference is {{{d=2}}}, the second is 

{{{a[2]=12+2(2-1)}}}
{{{a[2]=12+2}}}
{{{a[2]=14}}}

and third term is 
{{{a[3]=14+2}}}
{{{a[3]=16}}}

the first four terms of the sequence are:  {{{12}}},{{{14}}},{{{16}}}, {{{18}}}



3. In a geometric sequence, if {{{a[3] = -5}}} and {{{a[6] = 40}}}, determine {{{a[1]}}}, {{{r}}}, and {{{a[n]}}}.
Then write the first three terms of the sequence.

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

use  {{{a[3] = -5}}}

{{{-5=a[1]*r^(3-1)}}}
{{{-5=a[1]*r^2}}}..........solve for {{{a[1]}}}
{{{a[1]=-5/r^2}}}................eq.1

use  {{{a[6] = 40}}}

{{{40=a[1]*r^(6-1)}}}
{{{40=a[1]*r^5}}}..........solve for {{{a[1]}}}
{{{a[1]=40/r^5}}}................eq.2

from eq.1 and eq.2 we have
{{{-5/r^2=40/r^5}}}.......solve for {{{r}}}
{{{-5/40=r^2/r^5}}}
{{{-5/40=1/r^3}}}.......cross multiply

{{{-5r^3=1*40}}}
{{{r^3=40/-5}}}
{{{r^3=-8}}}
{{{r^3=-2^3}}}
{{{r=-2}}}

go to

{{{a[1]=-5/r^2}}}................eq.1, plug in {{{r}}}

{{{a[1]=-5/(-2)^2}}}

{{{a[1]=-5/4}}}

so, nth term formula is

{{{a[n]=(-5/4)*(-2)^(n-1)}}}

to write the first three terms of the sequence, we need to find second term
{{{n=2}}}
{{{a[2]=(-5/4)*(-2)^(2-1)}}}
{{{a[2]=(-5/4)*(-2)^1}}}
{{{a[2]=(5/4)*2}}}
{{{a[2]=(5/cross(4)2)*cross(2)}}}
{{{a[2]=5/2}}}

so, the first three terms of the sequence are: {{{-5/4}}},{{{5/2}}}, {{{ -5}}}