Question

Find an A.P. in which the sum of any number of terms is always three times the squared number of these terms.​

Answer 1

Answer:

Thus,

$\sf ◆S_n=3m^2\\ \\ \sf ◆ \: For \: n =1, S_1=3×1^2=3\\ \\ \sf ◆\sf \: For \: N =2, S_2=3×2^2=12\\ \\ \sf ◆ \: For \: N =3, S_3=3×3^3=27$

And So on,

$\sf \: \therefore S_1=A_1=3 \\ \\ \sf \: ◆ \: \: A_2=S_2-S_1=12-3=9\\ \\ \sf ◆ \: \: A_3=S_3-S_2=27-12=15$

◆ And So On

Thus The AP is 3,9,27

Answer 2

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

Given that,

• An A.P. series in which the sum of any number of terms is always three times the squared number of these terms.

Let assume that

• First term of an AP = a

• Common difference of an AP = d

• Number of terms = n

So, According to statement

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

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.

So, using this, we get

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

$\rm :\longmapsto\:2a + (n - 1)d = 6n$

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

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

So, on comparing, we get

$\rm \implies\:\boxed{ \tt{ \: d \: = \: 6 \: }}$

and

$\rm :\longmapsto\:2a - d = 0$

$\rm :\longmapsto\:2a - 6 = 0$

$\rm :\longmapsto\:2a = 6$

$\rm \implies\:\boxed{ \tt{ \: a \: = \: 3 \: }}$

So, Required AP series is

$\red{\rm :\longmapsto\:3, \: 9, \: 15, \: 21, - - - }$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

↝ nᵗʰ term of an arithmetic sequence is,

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

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.