Question

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This is a question of class 11th .
Chapter- Progression and Series.

Ques- Sum of infinite terms in a G.P. with common difference less than |1| .
Derive the expression.

Answer by mods, bs and serious users only.​

Answer 1

So we're asked to derive the expression for the sum of infinite terms in a GP with common ratio (not common difference) having magnitude less than 1, i.e., $\sf{|r|\,\textless\ 1.}$

First we derive out the expression for the sum of $\sf{n}$ terms in a GP.

Since the $\sf{n^{th}}$ term of a GP of first term a and common ratio r is $\sf{ar^{n-1},}$ let S be the sum of first n terms of the GP such that,

$\longrightarrow\sf{S=a+ar+ar^2+\,\dots\,+ar^{n-1}}$

Taking a as common in RHS,

$\longrightarrow\sf{S=a\left(1+r+r^2+\,\dots\,+r^{n-1}\right)}$

$\longrightarrow\sf{\dfrac{S}{a}=1+r+r^2+\,\dots\,+r^{n-1}\quad\quad\dots(1)}$

$\longrightarrow\sf{\dfrac{S}{a}-1=r+r^2+\,\dots\,+r^{n-1}}$

Taking r as common in RHS,

$\longrightarrow\sf{\dfrac{S-a}{a}=r\left(1+r+r^2+\,\dots\,+r^{n-2}\right)}$

$\longrightarrow\sf{\dfrac{S-a}{a}=r\left(1+r+r^2+\,\dots\,+r^{n-2}+r^{n-1}-r^{n-1}\right)}$

From (1),

$\longrightarrow\sf{\dfrac{S-a}{a}=r\left(\dfrac{S}{a}-r^{n-1}\right)}$

$\longrightarrow\sf{\dfrac{S-a}{a}=\dfrac{Sr}{a}-r^n}$

$\longrightarrow\sf{\dfrac{Sr-S+a}{a}=r^n}$

$\longrightarrow\sf{S(r-1)+a=ar^n}$

$\longrightarrow\sf{S(r-1)=ar^n-a}$

$\longrightarrow\sf{S=\dfrac{a(r^n-1)}{r-1}}$

For $\sf{|r|\,\textless\ 1,}$ the sum can be written without changing its value as,

$\longrightarrow\sf{S=\dfrac{a(1-r^n)}{1-r}}$

Sum of infinite terms is obtained for n tends to infinity,

$\displaystyle\longrightarrow\sf{S=\lim_{n\to\infty}\dfrac{a(1-r^n)}{1-r}}$

Since $\sf{|r|\,\textless\ 0,\ n\to\infty\ \implies\ r^n\to0.}$ Therefore,

$\displaystyle\longrightarrow\sf{\underline{\underline{S=\dfrac{a}{1-r}}}}$

Hence Derived!

Answer 2

So we're asked to derive the expression for the sum of infinite terms in a GP with common ratio (not common difference) having magnitude less than 1, i.e., \sf{|r|\,\textless\ 1.}∣r∣\textless 1.

First we derive out the expression for the sum of \sf{n}n terms in a GP.

Since the \sf{n^{th}}n

th

term of a GP of first term a and common ratio r is \sf{ar^{n-1},}ar

n−1

, let S be the sum of first n terms of the GP such that,

\longrightarrow\sf{S=a+ar+ar^2+\,\dots\,+ar^{n-1}}⟶S=a+ar+ar

2

+…+ar

n−1

Taking a as common in RHS,

\longrightarrow\sf{S=a\left(1+r+r^2+\,\dots\,+r^{n-1}\right)}⟶S=a(1+r+r

2

+…+r

n−1

)

\longrightarrow\sf{\dfrac{S}{a}=1+r+r^2+\,\dots\,+r^{n-1}\quad\quad\dots(1)}⟶

a

S

=1+r+r

2

+…+r

n−1

…(1)

\longrightarrow\sf{\dfrac{S}{a}-1=r+r^2+\,\dots\,+r^{n-1}}⟶

a

S

−1=r+r

2

+…+r

n−1

Taking r as common in RHS,

\longrightarrow\sf{\dfrac{S-a}{a}=r\left(1+r+r^2+\,\dots\,+r^{n-2}\right)}⟶

a

S−a

=r(1+r+r

2

+…+r

n−2

)

\longrightarrow\sf{\dfrac{S-a}{a}=r\left(1+r+r^2+\,\dots\,+r^{n-2}+r^{n-1}-r^{n-1}\right)}⟶

a

S−a

=r(1+r+r

2

+…+r

n−2

+r

n−1

−r

n−1

)

From (1),

\longrightarrow\sf{\dfrac{S-a}{a}=r\left(\dfrac{S}{a}-r^{n-1}\right)}⟶

a

S−a

=r(

a

S

−r

n−1

)

\longrightarrow\sf{\dfrac{S-a}{a}=\dfrac{Sr}{a}-r^n}⟶

a

S−a

=

a

Sr

−r

n

\longrightarrow\sf{\dfrac{Sr-S+a}{a}=r^n}⟶

a

Sr−S+a

=r

n

\longrightarrow\sf{S(r-1)+a=ar^n}⟶S(r−1)+a=ar

n

\longrightarrow\sf{S(r-1)=ar^n-a}⟶S(r−1)=ar

n

−a

\longrightarrow\sf{S=\dfrac{a(r^n-1)}{r-1}}⟶S=

r−1

a(r

n

−1)

For \sf{|r|\,\textless\ 1,}∣r∣\textless 1, the sum can be written without changing its value as,

\longrightarrow\sf{S=\dfrac{a(1-r^n)}{1-r}}⟶S=

1−r

a(1−r

n

)

Sum of infinite terms is obtained for n tends to infinity,

\displaystyle\longrightarrow\sf{S=\lim_{n\to\infty}\dfrac{a(1-r^n)}{1-r}}⟶S=

n→∞

lim

1−r

a(1−r

n

)

Since \sf{|r|\,\textless\ 0,\ n\to\infty\ \implies\ r^n\to0.}∣r∣\textless 0, n→∞ ⟹ r

n

→0. Therefore,

\displaystyle\longrightarrow\sf{\underline{\underline{S=\dfrac{a}{1-r}}}}⟶

S=

1−r

a

Hence Derived!