Question

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

Answer 1

`+-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 2

`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 3

`"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`