The second term of a geometric progressionis 6.5 and the sum to infini

1.	The second term of a geometric progressionis 6.5 and the sum to infinity is 26.Find the common ratio and the first term.
Answer» $\large\underline{\bf{Solution-}}$ 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

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

Answer»

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

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