Math, asked by aayat403, 27 days ago

The second term of a geometric progression
is 6.5 and the sum to infinity is 26.
Find the common ratio and the first term.​

Answers

Answered by mathdude500
4

\large\underline{\bf{Solution-}}

We know that,.

↝ nᵗʰ term of an geometric sequence is,

\bf :\longmapsto\:a_n =  {ar}^{n - 1}

Wʜᴇʀᴇ,

  • aₙ is the nᵗʰ term.

  • a is the first term of the sequence.

  • n is the no. of terms.

  • r is the common ratio.

Tʜᴜs,

↝ second term is,

\bf :\longmapsto\:a_2 =  {ar}^{2 - 1} = ar

As second term is 6.5

\bf\implies \:ar = 6.5 -  -  -  - (1)

Also,

We know that,

↝ Sum of infinite geometric sequence is,

\rm :\longmapsto\:S_ \infty  = \dfrac{a}{1 - r}, \: provided \: that \:  |r| < 1

According to statement,

\rm :\longmapsto\:S_ \infty  = 26

\rm :\implies\:\dfrac{a}{1 - r} = 26

\bf\implies \:a = 26(1 - r) -  -  - (2)

On substituting the value of a in equation (1), we get

\rm :\longmapsto\:26(1 - r)r = 6.5

\rm :\longmapsto\:26(1 - r)r =  \dfrac{13}{2}

\rm :\longmapsto\:2(1 - r)r =  \dfrac{1}{2}

\rm :\longmapsto\:4(1 - r)r =  1

\rm :\longmapsto\:4r -  {4r}^{2} = 1

\rm :\longmapsto\: {4r}^{2} - 4r + 1 = 0

\rm :\longmapsto\: {(2r - 1)}^{2}  = 0

\rm :\longmapsto\:2r - 1 = 0

\bf\implies \:r = \dfrac{1}{2}  -  - (3)

On substituting the value of r in equation (1) we get

\bf\implies \:a = 26(1 - \dfrac{1}{2} )

\bf\implies \:a = 13

Additional Information :-

1. If a, b, c are in GP then

\rm :\longmapsto\: {b}^{2} = 4ac

2. Sum of n terms of GP series is

\begin{gathered}\begin{gathered}\bf\: S_n = \begin{cases} &\sf{a\bigg(\dfrac{ {r}^{n} - 1 }{r - 1} \bigg)}, \:  \: if \: r \ne \: 1\\ \\  &\sf{na, \:  \: if \: r \:  =  \: 1} \end{cases}\end{gathered}\end{gathered}

Wʜᴇʀᴇ,

  • Sₙ is the sum of first n terms.

  • a is the first term of the sequence.

  • n is the no. of terms.

  • r is the common ratio

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