Geometric Sequence

By setting $a_1=a$ and $a_k=aq^{k-1}$ for $k\in \mathbb N$ , we have

$a\cdot aq=aq^9 \Rightarrow a=q^8$

$a+aq^8=20 \Rightarrow a^2+a-20=0 \Rightarrow a=4, \quad q^4=2$

Therefore

$\quad \left(a+aq^2+aq^4+aq^6+aq^8\right)\left(a-aq^2+aq^4-aq^6+aq^8\right)$

$=a^2 \left(1+q^2+q^4+q^6+q^8\right)\left(1-q^2+q^4-q^6+q^8\right)$

$=a^2\frac{1-q^{10}}{1-q^2}\frac{1+q^{10}}{1+q^2}$

$=a^2\frac{1-q^{20}}{1-q^4}$

$=a^2\frac{1-(q^4)^{5}}{1-q^4}$

$=16\frac{1-32}{1-2}=496$

Hey yo,

As for this, stated that the sequence is always +,

therefore as a(1).a(2) = a(10),

a(ar) = ar^(9)

a = r^8(1st)

as for a(1) + a(9) = 20

a + ar^8 = 20(2nd), by substituting (1st) into (2nd),

a + a^(2) = 20

a^(2) + a - 20 = 0

(a -4) ( a + 5) = 0

a = 4 or a = -5, as stated the sequence is always > 0 ( + ), r^8 = 4 , r^8 = 2^2 , r = (2^2)^(1/8) = 2^(1/4),

therefore a = 4,

a(1) =4 , a(3) = 4(2^(1/4))^2=4(2)^(1/2), a(5) = 4(2) =8 , a(7) = 4(2^(1/4))^6 = 4(2)^(3/2)

a(9) = ar^8 = 4.4 = 16,

so just by substituting into ( a1+a3+a5+a7+a9)(a1-a3+a5-a7+a9),

= ( 4 + 4(2)^(12) + 8 + 4(2)^(3/2) + 16) ( 4 - 4(2)^(1/2) + 8 - 4(2)^(3/2) + 16)

= ( 28 + 4(2)^0.5 + 4(2)^(1.5) ) ( 28 - 4(2)^(0.5) - 4(2)^(1.5) )

= 496...

thanks...

Nice prob.