Exams Joint Entrance Examination Mathematics Sequences and Series Question 174 Question 174 of 519 Hard NUMERICAL
Let { a k } \{a_k\} { a k } and { b k } \{b_k\} { b k } , k ∈ N k\in\mathbb{N} k ∈ N , be two G.P.s with common ratios r 1 r_1 r 1 and r 2 r_2 r 2 respectively, such that a 1 = b 1 = 4 a_1=b_1=4 a 1 = b 1 = 4 and r 1 < r 2 r_1<r_2 r 1 < r 2 . Let c k = a k + b k c_k=a_k+b_k c k = a k + b k . If c 2 = 5 c_2=5 c 2 = 5 and c 3 = 13 4 c_3=\dfrac{13}{4} c 3 = 4 13 , then ∑ k = 1 ∞ c k − ( 12 a 6 + 8 b 4 ) \displaystyle\sum_{k=1}^{\infty}c_k - (12a_6+8b_4) k = 1 ∑ ∞ c k − ( 12 a 6 + 8 b 4 ) is equal to \_\_\_\_\_\_\_\_.
Answer: 9
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c 2 = 4 r 1 + 4 r 2 = 5 ⇒ r 1 + r 2 = 5 4 c_2 = 4r_1+4r_2 = 5 \Rightarrow r_1+r_2=\dfrac54 c 2 = 4 r 1 + 4 r 2 = 5 ⇒ r 1 + r 2 = 4 5 ; c 3 = 4 r 1 2 + 4 r 2 2 = 13 4 ⇒ r 1 2 + r 2 2 = 13 16 c_3 = 4r_1^2+4r_2^2=\dfrac{13}{4} \Rightarrow r_1^2+r_2^2=\dfrac{13}{16} c 3 = 4 r 1 2 + 4 r 2 2 = 4 13 ⇒ r 1 2 + r 2 2 = 16 13 .
\[r_1r_2 = \frac{(r_1+r_2)^2-(r_1^2+r_2^2)}{2} = \frac{25/16-13/16}{2} = \frac38.\]
Solving t 2 − 5 4 t + 3 8 = 0 t^2-\tfrac54t+\tfrac38=0 t 2 − 4 5 t + 8 3 = 0 : t = 3 4 t=\tfrac34 t = 4 3 or 1 2 \tfrac12 2 1 ; since r 1 < r 2 r_1<r_2 r 1 < r 2 , r 1 = 1 2 r_1=\tfrac12 r 1 = 2 1 , r 2 = 3 4 r_2=\tfrac34 r 2 = 4 3 .
\[\sum_{k=1}^{\infty}c_k = \frac{4}{1-1/2}+\frac{4}{1-3/4} = 8+16 = 24.\]
\[a_6 = 4\left(\frac12\right)^5=\frac18,\quad b_4=4\left(\frac34\right)^3=\frac{27}{16} \implies 12a_6+8b_4 = \frac32+\frac{27}{2}=15.\]
24 − 15 = 9. 24-15 = 9. 24 − 15 = 9. Answer the question to unlock