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Question

Mathematics Question on Sequence and series

Let ana_n be the nthn^{th} term of a G.P. of positive terms. If x=n=1100a2n+1=200x=\displaystyle \sum_{n=1}^{100}a_{2n+1}=200 and x=n=1100a2n=100x=\displaystyle \sum_{n=1}^{100}a_{2n}=100, x=n=1200anx=\displaystyle \sum_{n=1}^{200}a_n is equal to :

A

300300

B

175175

C

225225

D

150150

Answer

150150

Explanation

Solution

n=1100a2n+1=200a3+a5+a7+....+a201=200\sum\limits^{100}_{n = 1} a_{2n+1} = 200 \Rightarrow a_{3} + a_{5} + a_{7} + .... + a_{201} = 200
ar2(r2001)(r21)=200\Rightarrow ar^{2} \frac{\left(r^{200}-1\right)}{\left(r^{2}-1\right)} = 200
n=1100a2n=100a2+a4+a6+...+a200=100\sum\limits^{100}_{n = 1} a_{2n} = 100 \Rightarrow a_{2} + a_{4} + a_{6} + ... + a_{200} = 100
ar(r2001)(r21)=100\Rightarrow \frac{ar\left(r^{200}-1\right)}{\left(r^{2}-1\right)} = 100
On dividing r=2r = 2
on adding a2+a3+a4+a5+...+a200+a201=300a_{2} + a_{3} + a_{4} + a_{5} + ... +a_{200} + a_{201} = 300
r(a1+a2+a3+....+a200)=300\Rightarrow r\left(a_{1} + a_{2} + a_{3} + .... + a_{200}\right) = 300
n=1100an=150\Rightarrow \sum\limits^{100}_{n = 1} a_{n} = 150