Question

If sum of 1st nth terms of AP is Sn=3n(n+1)
find AP​

Answer 1

Step-by-step explanation:

Given:-

Sum of first n terms of an AP Sn = 3n(n+1)

To find:-

Find the Arithmetic Progression?

Solution:-

Given that

Sum of first 'n' terms of an AP is Sn = 3n(n+1)

Put n = 1 then

=>S1 = 3(1)(1+1)

=>S1 = 3(2)

=>S1 = 6

First term of the AP = 6

Put n = 2 then

=>S2 = 3(2)(2+1)

=>S2 = 6(3)

=>S2 = 18

=> Sum of first two terms = 18

=>First term + Second term = 18

=>6+ Second term = 18

=>Second term = 18-6

Second term = 12

we have

First term (t1) = 6

Second term (t2) = 12

Common difference (d) = t2 - t1

=>d = 12-6

=>d = 6

Common difference of the AP = 6

Now The general form of an AP is

t1 , t1+d , t1+2d,..

t1 = 6

t1+d = 6+6 = 12

t1+2d = 6+2(6)=6+12=18

The AP : 6,12,18,...

Answer :-

The required Arithmetic Progression is

6 , 12 , 18 , 24, 30 ,...

Check:-

The AP : 6,12,18,24,...

t1 = 6

d = 12-6 = 6

Sum of diary n terms of an AP

Sn =(n/2)[2t1+(n-1)d]

=>Sn = (n/2)[2(6)+(n-1)6]

=>Sn = (n/2)(12+6n-6)

=>Sn = (n/2)(6n+6)

=>Sn = 6(n)(n+1)/2

=>Sn = 3n(n+1)

Verified the given relation.

Used formulae:-

1.If t1 is the first term and d is the common difference then d = tn - t(n-1).

2. Sum of first n terms of an AP= Sn =

(n/2)[2t1+(n-1)d]

n is the number of terms in an AP

Answer 2

$\begin{gathered}\begin{gathered}\bf \: Given \:  - \begin{cases} &\sf{S_{n} = 3n(n + 1)} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To\: find - \begin{cases} &\sf{ \: \: AP \: series}  \end{cases}\end{gathered}\end{gathered}$

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

• Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

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

Wʜᴇʀᴇ,

• Sₙ is the Sum of first n term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

Cᴀʟᴄᴜʟᴀᴛɪᴏɴ

$\rm :\longmapsto\:S_{n} = 3n(n + 1)$

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

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

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

On comparing, we get

$\rm :\longmapsto\:d = 6 \: \: \: and \: \: \: 2a - d = 6$

$\rm :\implies\:2a - 6 = 6\rm :\implies\:a = 6$

Hence,

• The required series is 6, 12, 18, 24, __

Additional Information

• ↝ 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.