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AGP Series 1+4x+7x^2+... = 35/16: Find x | JEE

JEE Maths question with a full step-by-step solution.

Question
If the sum to infinity of 1+4x+7x2+10x3+⋯1+4x+7x^2+10x^3+\cdots is 3516\frac{35}{16}, where ∣x∣<1|x|<1, then xx equals
A197\frac{19}{7}
B15\frac{1}{5}correct
C14\frac{1}{4}
DNone of these
Solution
Step 1: Let S=1+4x+7x2+10x3+⋯S=1+4x+7x^2+10x^3+\cdots and form S−xSS-xS:
S−xS=1+3x+3x2+3x3+⋯ ,S-xS=1+3x+3x^2+3x^3+\cdots,
 where the tail after 11 is a G.P. with first term 3x3x and ratio xx. Step 2:
S(1−x)=1+3x1−x=1+2x1−x ⇒ S=1+2x(1−x)2.S(1-x)=1+\frac{3x}{1-x}=\frac{1+2x}{1-x}\ \Rightarrow\ S=\frac{1+2x}{(1-x)^2}.
 Step 3: Put S=3516S=\dfrac{35}{16}:
16(1+2x)=35(1−x)2 ⇒ 35x2−102x+19=0.16(1+2x)=35(1-x)^2\ \Rightarrow\ 35x^2-102x+19=0.
 Step 4: Factor:
(5x−1)(7x−19)=0 ⇒ x=15 or 197.(5x-1)(7x-19)=0\ \Rightarrow\ x=\frac15\ \text{or}\ \frac{19}{7}.
 Since ∣x∣<1|x|<1, take x=15.x=\dfrac15. Correct answer: (2)
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