Hi

This is combination of geometric and arithmetic sequences.

n>0, min n = 1.

\(a_2 = 10 + \frac{1}{2}a_1\)

\(a_3 = 10 + \frac{1}{2}a_2 = 10 + \frac{1}{2}(10 + \frac{1}{2}a_1) = 10 + 5 + \frac{1}{4}a_1\)

\(a_4 = 10 + \frac{1}{2}a_3 = 10 + \frac{1}{2}(10 + 5 + \frac{1}{4}a_1) = 10 + 5 + \frac{5}{2} + \frac{1}{8}a_1\)

\(a_5 = 10 + \frac{1}{2}a_4 = 10 + 5 + \frac{5}{2} + \frac{5}{4} + \frac{1}{16}a_1\)

...

\(a_n = 10 + 5 + \frac{5}{2} + \frac{5}{4} + ... + \frac{5}{2^{n-3}} + \frac{1}{2}^{n-1}a_1\)

When n=1000 our fraction \(\frac{1}{2^{999}}\) is close to 0.

\(5 + \frac{5}{2} + \frac{5}{4} + ... + \frac{5}{2^{997}}\) We can apply logic of infinite geometric series where |r|<1 because our n is quite big (1000).

\(S = \frac{5}{1-1/2} = 5*2 = 10\)

\(a_{1000} ≈ 10 + 10 + 0 = 20\)

Answer C

Hope this helps

Regards

Hi,


A point before the solution..
The Q is flawed in that there is no value of \(A_1\) given.
Solution..
\(A_1=10, A_2=10+0.5*10=10+5, A_3=10+5+0.5*5=10+5+0.25=10+10/2+10/4+.....\)
So 1000 can be taken as infinite series..
Ans =\(\frac{a}{(1-r)}=10/(1-1/2)=10/(1/2)=20\)

Can you please explain to me how you got to the 10/(1-1/2) part in the last equation? I can not seem to trace the origin of the 1/2 part and why that expression is the divisor of 10.
Thank you.

Given : A(n+1) = 10 + A(n)/2

A(2) = 10+ A(1)/2
A(3) = 10+ A(2)/2 = 15+A(1)/4
A(4) = 10+ A(3)/2 = 17.5+A(1)/8
A(5) = 10+ A(4)/2 = 18.75+A(1)/16
A(6) = 10+A(5)/2 = 19.375+A(1)/32
a(7) = 19.6875 + a(1)/64
so the second term for A(1000) somewhat 19.XxXXXXXx+A(1)/2^999 , the second part can be ignored closet answer would be 20.
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I see now, throughout all equations you utilize A(1) and the end result is neglectable. Thanks a million!

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