Question

Please solve this question.​

Answer 1

Answer:

I think option c,

Step-by-step explanation:

please mark me brainliest

Answer 2

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

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

Thus, According to statement,

$\rm :\longmapsto\:S_{2n} \: = \: 3S_n$

$\rm :\longmapsto\:\dfrac{2n}{2} \bigg(2a + (2n - 1)d \bigg) = \dfrac{3n}{2} \bigg(2a + (n - 1)d\bigg)$

$\rm :\longmapsto\:2 \bigg(2a + (2n - 1)d \bigg) = 3 \bigg(2a + (n - 1)d\bigg)$

$\rm :\longmapsto\:4a + (4n - 2)d = 6a + (3n - 3)d$

$\rm :\longmapsto\:4a + 4nd - 2d = 6a + 3nd - 3d$

$\rm \implies\:\boxed{ \tt{ \: 2a = nd + d \: }} - - - (1)$

Now, Consider

$\rm :\longmapsto\:\dfrac{S_{3n}}{S_n}$

$\rm \:  =  \:\dfrac{\dfrac{3n}{2} \bigg(2a + (3n - 1)d\bigg) }{\dfrac{n}{2} \bigg(2a + (n - 1)d \bigg) }$

On substituting the values of 2a, we get

$\rm \:  =  \:\dfrac{3 \bigg(nd + d+ 3nd - d\bigg) }{ \bigg(nd + d + nd - d \bigg) }$

$\rm \:  =  \:\dfrac{3 \bigg(4nd \bigg) }{ \bigg(2nd \bigg) }$

$\rm \:  =  \:6$

Hence,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{S_{3n}}{S_n} = 6 \: }}}$