What is the 7th term of the geometric sequence where a1 and a4 64?

Given a geometric sequence with the first term a1 and the common ratio r , the nth (or general) term is given by
an=a1⋅rn−1 .

Example 1:

Find the 6th term in the geometric sequence 3,12,48,... .

a1=3,  r=123=4a6=3⋅46−1=3⋅45=3072

Example 2:

Find the 7th term for the geometric sequence in which a2=24 and a5=3 .

Substitute 24 for a2 and 3 for a5 in the formula

an=a1⋅rn−1 .

a2=a1⋅r2−1→24=a1ra5=a1⋅r5−1→    3=a1r4

Solve the firstequation for a1 : a1=24r

Substitute this expression for a1 in the second equation and solve for r .

3=24r⋅r43=24r318=r3  so  r=12

Substitute for r in the first equation and solve for a1 .

24=a1(12)48=a1

Now use the formula to find a7 .

a7=48(12)7−1=48⋅164=34

See also: sigma notation of a series and nth term of a arithmetic sequence

Precalculus Sequences Geometric Sequences

3 Answers

Ratnaker Mehta

Aug 2, 2018

# +-500#.

Explanation:

In the Usual Notation,#a_n=a_1r^(n-1), n in NN#.

Given that,#a_6=-100 and a_4=-4#.

#rArr a_1r^5=-100 and a_1r^3=-4#.

#:. (a_1r^5)/(a_1r^3)=(-100)/(-4)=25#.

#:. r^2=25.#

#:. r=+-5#.

#r=-5, &, a_4=-4 rArr a_1=-4/r^3=(-4)/(-5)^3=4/5^3#.

Then,#a_6=a_1r^5=(4/5^3)*(-5)^5=-100#.

Hence, in this case,#a_7=a_1r^6=(4/5^3)*(-5)^6=500#.

In case,#r=+5, then, a_1=-4/5^3, &, a_6=-4/5^3*5^5=-100#

Then,#a_7=a_6*r=(-100)(+5)=-500#.

Answer link

Lucy

Aug 2, 2018

#T_7=+-500#

Explanation:

We know that any term in a geometric sequence can be described as#T_n=ar^(n-1)#
where
#a#is the first term
#n#is the nth term
#r#is the ratio between 2 adjacent terms

If we know that#T_4=-4#and#T_6=-100#and#T_n=ar^(n-1)#, we can solve to find#a#and#r#

#T_4=ar^(4-1)=-4#
#T_4=ar^3=-4#---- (1)

#T_6=ar^(6-1)=-100#
#T_6=ar^5=-100#---- (2)

#((2))/((1))#

#(ar^5)/(ar^3)=-100/-4#

#r^2=25#

#r=+-5#

If we know that#r=5#, then subbing#r=5#back into (1)
#a(5)^3=-4#
#125a=-4#
#a=-4/125#

To test if it is correct, sub#a=-4/125#into (2)

#LHS#
#=-4/125times5^5#
#=-4/125times3125#
#=-100#
#=RHS#

If we know that#r=-5#, then subbing#r=-5#back into (1)
#a(-5)^3=-4#
#-125a=-4#
#a=4/125#

To test if it is correct, sub#a=4/125#into (2)

#LHS#
#=4/125times(-5)^5#
#=4/125times-3125#
#=-100#
#=RHS#

Therefore, we know that#r=+-5#and#a=+-4/125#
To find#T_7#,
#T_7=+-4/125times5^(7-1)=+-4/125times5^6=+-500#

Answer link

maganbhai P.

Aug 2, 2018

#"The "7^(th)"term of geometric sequence is:"#

#a_7=500or -500#

Explanation:

We know that,

#color(green)(n^(th) "term of the Geometric sequence is :"#

#color(green)(a_n=a_1(r)^(n-1)#

#where ,color(green)(a_1="first term" and r="common ratio."#

We have,

#a_4=-4color(white)(;;;;.............;;;;)and a_6=-100#

#:.a_4=a_1(r)^(4-1)=-4 color(white)(;;;)and a_6=a_1(r)^(6- 1)=-100#

#:.a_1r^3=-4....to(1)and a_1r^5=-100...to(2)#

#eqn. (2)=>a_1r^3*r^2=-100#

#=>(-4)r^2=-100...touse ,eqn(1)#

#=>r^2=(-100)/(-4)=25#

#color(red)(=>r+-5#

Now ,

#(i)"for " color(red)( r=5#

#7^(th)term=a_7=a_1(r)^(7-1)#

#=>a_7=a_1r^6=a_1r^5*r^1#

#=>a_7=(-100)(+5)to[because eqn.(2) and color(red)(r=5)]#

#=>color(blue)(a_7=-500#

#(i)"for " color(red)( r=-5#

#7^(th)term=a_7=a_1(r)^(7-1)#

#=>a_7=a_1r^6=a_1r^5*r^1#

#=>a_7=(-100)(-5)to[because eqn.(2) and color(red)(r=-5)]#

#=>color(blue)(a_7=500#

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