Question

Solve the question number 8 and 10.
Please explain.
​

Answer 1

$\green{\large\underline{\sf{Solution-10}}}$

Given that,

↝ pᵗʰ, qᵗʰ, rᵗʰ of a GP are in GP.

Let assume that first term of the GP series is a and Common ratio is R.

So,

We know that,

↝ nᵗʰ term of an geometric sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\: {R}^{n - 1} }}}}}} \\ \end{gathered}$

So,

$\rm \implies\: {p}^{th} \: term \: = {aR}^{p - 1}$

$\rm \implies\: {q}^{th} \: term \: = {aR}^{q - 1}$

$\rm \implies\: {r}^{th} \: term \: = {aR}^{r - 1}$

Now,

It is given that,

↝ pᵗʰ, qᵗʰ, rᵗʰ of a GP are in GP

So, it means common ratio between the consecutive terms is same.

$\rm \implies\:\dfrac{ {aR}^{q - 1} }{{aR}^{p - 1}} = \dfrac{{aR}^{r - 1}}{{aR}^{q - 1}}$

$\rm \implies\:\dfrac{ {R}^{q - 1} }{{R}^{p - 1}} = \dfrac{{R}^{r - 1}}{{R}^{q - 1}}$

$\rm \implies\: {R}^{q - 1 - (p - 1)} = {R}^{r - 1 - (q - 1)}$

$\rm \implies\: {R}^{q - 1 - p + 1} = {R}^{r - 1 - q + 1}$

$\rm \implies\: {R}^{q- p} = {R}^{r - q}$

$\bf\implies \:q - p = r - q$

$\rm \implies\:\boxed{ \tt{ \: \: p,q,r \: are \: in \: AP \: \: }}$

• Hence, Option (a) is correct

$\red{\large\underline{\sf{Solution-8}}}$

Let assume that

$\rm :\longmapsto\:a_1,a_2,a_3, - - - ,a_n \: be \: an \: AP \: series \: with \: common \: difference \: d$

So,

↝ mᵗʰ term from the beginning is given by

$\rm \implies\:\boxed{ \tt{ \: a_m = a + (m - 1)d \: }} - - - (1)$

Now,

↝ mᵗʰ term from the end is given by

$\rm \implies\:\boxed{ \tt{ \: a_m = a_n - (m - 1)d \: }} - - - (2)$

Thus,

Sum of mᵗʰ term from beginning and mᵗʰ term from end is

$\rm \:  =  \:a + (m - 1)d + a_n - (m - 1)d$

$\rm \:  =  \:a + a_n$

= sum of first term and last term.

• Hence, Option (c) is correct.