Question

The term of a GP are distinct if the second term is 4 and the sum of its third & fourth terms is 8 the sixth term will be?

Answer 1

ar^1=4

ar^2+ar^3=8

so a=4/r

now putting in second equation

(4/r)r²+(4/r)r³=8

4r+4r²=8

4r²+4r-8=0

r²+r-2=0

r²+2-r-2=0

r(r+2)-1(r+2)=0

so r=1 or r=-2.

a=4 or a=-2

now sixth term ar5=4 or 64

hope it helped u

plz mark brainliest

Answer 2

The sixth term will be either 4 or 64.

Step-by-step explanation:

The nth term of a GP is

$a_n=ar^{n-1}$

where, a is first term and r is common ratio.

The second term is 4.

$a_2=ar^{2-1}$

$4=ar$                .... (1)

The sum of its third & fourth terms is 8.

$a_3+a_4=ar^{3-1}+ar^{4-1}$

$8=ar^{2}+ar^{3}$       ... (2)

Using (1) and (2) we get

$8=4r+4r^2$

Divide both sides by 4.

$2=r+r^2$

$0=r^2+r-2$

Spilling the middle term we get

$r^2+2r-r-2=0$

$r(r+2)-(r+2)=0$

$(r+2)(r-1)=0$

Using zero product property we get

$r=-2,1$

The 6th term of a GP is

$a_6=ar^{6-1}=ar^5$

If r=-2, then

$4=a(-2)\Rightarrow a=-2$

$a_6=(-2)(-2)^5=64$

If r=1, then

$4=a(1)\Rightarrow a=4$

$a_6=(4)(1)^5=4$

Therefore, the sixth term will be either 4 or 64.

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