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

3 Answers
Aug 2, 2018

+-500±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

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

"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