<span>30 UK£ is the answer
</span>
Third term = t3 = ar^2 = 444 eq. (1)
Seventh term = t7 = ar^6 = 7104 eq. (2)
By solving (1) and (2) we get,
ar^2 = 444
=> a = 444 / r^2 eq. (3)
And ar^6 = 7104
(444/r^2)r^6 = 7104
444 r^4 = 7104
r^4 = 7104/444
= 16
r2 = 4
r = 2
Substitute r value in (3)
a = 444 / r^2
= 444 / 2^2
= 444 / 4
= 111
Therefore a = 111 and r = 2
Therefore t6 = ar^5
= 111(2)^5
= 111(32)
= 3552.
<span>Therefore the 6th term in the geometric series is 3552.</span>
B = (2x+3)(4x^2-6x+9)-2(4^3-1)
B = 8x^3-99
Hope it helps : )
there are (1/6)/(1/3) one-third in one sixth
Step-by-step explanation:
that means
no. of one-third is 1/2
Answer:
if I have my calculations right it would be £120
