Question

pls tell the solution for this​

Answer 1

Answer:

Let a and r be the first term and common ratio here respectively.

We're given an infinite geometric progression, the sequence would be something like this :

\sf \: a + ar + {ar}^{2} + \dots \dots \dots \: + \inftya+ar+ar

2

+………+∞

According to the given condition,

\sf \: a + ar = 1 \longrightarrow(1)a+ar=1⟶(1)

Every term in the GP is twice the sum of it's successive terms,

\sf \: a = 2( ar + {ar}^{2} + \dots + {ar}^{n - 1} ) \longrightarrow \: (2)a=2(ar+ar

2

+⋯+ar

n−1

)⟶(2)

Now, the above sequence forms a GP with a common ratio r and first term 1, such that r < 1.

Sum of terms in a GP :

\boxed{ \boxed{ \sf \: S = \dfrac{a}{1 - {r}^{n} } }}

S=

1−r

n

a

Since, sum of successive terms is equal to the preceding term.

\begin{gathered}\longrightarrow \sf a = \dfrac{2ar}{1 - r} \\ \\ \longrightarrow \sf \: 1 = \dfrac{2r}{1 - r} \\ \\ \longrightarrow \sf \: 1 - r = 2r \\ \\ \longrightarrow \: \underline{\boxed{ \sf r = \dfrac{1}{3} }}\end{gathered}

⟶a=

1−r

2ar

⟶1=

1−r

2r

⟶1−r=2r

⟶

r=

3

1

Using condition (1),

\begin{gathered}\implies \sf \: a + \dfrac{a}{3} = 1 \\ \\ \implies \: \sf \: a = \dfrac{1}{ \big(1 + \dfrac{1}{3} \big)} \\ \\ \implies \boxed{ \boxed{ \sf \: a = \dfrac{3}{4} }}\end{gathered}

⟹a+

3

a

=1

⟹a=

(1+

3

1

)

1

⟹

a=

4

3

Step-by-step explanation:

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