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
How do you find the 7th term of the geometric sequence with the given terms a4 = -4, a6 = -100?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#
Answer link
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