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

3 Answers
Aug 2, 2018

Answer:

# +-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#.

Aug 2, 2018

Answer:

#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#

Aug 2, 2018

Answer:

#"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#