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Question

If the sum to infinity of the series $$1+4x+7x^2+10x^3+\cdots$$ is $$ \dfrac{35}{16}$$, then $$x=$$


A
15
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B
25
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C
37
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D
17
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Solution

The correct option is A $$\displaystyle \frac{1}{5}$$
Let$${ S }_{ \infty  }=1+4x+7{ x }^{ 2 }+10{ x }^{ 3 }+...\infty $$      $$...(1)$$

Now, multiply by $$x$$ throughout in eqution $$(1)$$; we get
$$x{ S }_{ \infty  }=x+4{ x }^{ 2 }+7{ x }^{ 3 }+10{ x }^{ 4 }+...\infty $$      $$...(2)$$
Subtracting $$(2)$$ from $$(1)$$; we get

$${S}_{ \infty  }-x{ S }_{ \infty  }=(1+4x+7{ x }^{ 2 }+10{ x }^{ 3 }+...\infty )-(x+4{ x }^{ 2 }+7{ x }^{ 3 }+10{ x }^{ 4 }+...\infty )$$

$$\Rightarrow(1-x){ S }_{ \infty  }=1+4x+7{ x }^{ 2 }+10{ x }^{ 3 }+...\infty -x-4{ x}^{ 2 }-7{ x }^{ 3 }-10{ x }^{ 4 }-...\infty $$
$$\Rightarrow (1-x){ S }_{ \infty  }=1+3x+3{ x }^{ 2 }+3{ x }^{ 3 }+...\infty $$
Notice that the series $$3x+3{ x }^{ 2 }+3{ x }^{ 3 }+...\infty $$ is geometric series with the first term $$a=3x$$ and the common ratio $$r=x$$.

Now, use the formula for the sum of an infinite geometric series.
$$\Rightarrow (1-x){ S }_{ \infty  }=1+\dfrac { 3x }{ (1-x) } $$, for$$\left| x \right| <1$$
$$\Rightarrow (1-x){ S }_{ \infty  }=\dfrac { 1+2x }{ (1-x) } $$, for$$\left| x \right| <1$$

Given that, $${ S }_{ \infty  }=\dfrac { 35 }{ 16 }$$, substitute for $${ S }_{ \infty  }$$ in the above equation; we get
$$(1-x)\dfrac { 35 }{ 16 } =\dfrac { 1+2x }{ (1-x) } $$
$$\Rightarrow35{ (1-x) }^{ 2 }=16(1+2x)$$

$$\Rightarrow 35(1-2x+{ x }^{ 2 })=16+32x$$
$$\Rightarrow 35{ x }^{ 2 }-102x+19=0$$

$$\Rightarrow (7x-19)(5x-1)=0$$
$$\Rightarrow x=\dfrac { 19 }{ 7 } $$ or $$x=\dfrac { 1 }{ 5 } $$

But $$x\neq \dfrac { 19 }{ 7 } $$, because for infinity series, $$\left| x \right| <1$$.

Therefore, $$x=\dfrac { 1 }{ 5 } $$.

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