Question

The sum of first n terms of an ap 3n2+6n find the nth term of this AP

Answer 1

Substitute n with 1 in Sn=3n^2+6n

S1=3(1)^2+6(1)

a=3+6 {S1=a}

a=9.

Substitute n with 2 in Sn=3n^2+6n

S2=3(2)^2+6(2)

S2=3(4)+12

S2=12+12

S2=24.

We know that

S2=a+a2

So,

S2=24

a+a2=24

9+a2=24

a2=15.

d=a2-a

d=15-9

d=6

So, with a=9 and d=6, we can find the nth term.

an=a+(n-1)d

an=9+(n-1)6

an=9+6n-6

an=6n+3.

Thus, nth term is (6n+3).

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Answer 2

Answer:

The nth term of the A.P is  6n + 3

Step-by-step explanation:

Given :  

Sn = 3n² + 6n  …………..(1)

Now , on Replacing n by (n –1) in eq (i),  

S(n – 1) = 3(n – 1)² + 6(n – 1)

nth term of the A.P., an = Sn – S(n – 1)

an = (3n² + 6n) – [3(n – 1)² + 6(n – 1)]

⇒ an =(3n² + 6n) – [3 (n² + 1 - 2n) + 6n - 6]

[(a + b)² = a² + b² - 2ab]

⇒ an =(3n² + 6n) – [3n² + 3 - 6n + 6n - 6]

⇒ an =(3n² + 6n) – [3n² -  3 ]

⇒ an = 3n² + 6n – 3n² + 3  

⇒ an =  6n  + 3  

Hence, the nth term of the A.P is  6n + 3

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