Sequences & SerieseasyFree

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)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Sequences & Series · easy
a, b, c, d are in increasing G.P. If the A.M. between a and b is 6 and the A.M. between c…
Sequences & Series · medium
Let a₁, a₂, …, a(4001) be in A.P. such that (1)/(a₁ a₂)+(1)/(a₂ a₃)+⋯+ 1 a(4000) a(4001)…
Sequences & Series · medium
If (p+q· 5^x)/(p-q· 5^x)=(q+r· 5^x)/(q-r· 5^x)=(r+s· 5^x)/(r-s· 5^x) , then p, q, r, s ar…
Sequences & Series · easy
Let a₁, a₂, … and b₁, b₂, … be arithmetic progressions with a₁=25 , b₁=75 and a(100)+b(10…

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath