Question

5. If S, the sum of first n terms of an A.P. is given by
s 3n2 - 4n, then find its nth term.​

Answer 1

Question:

• $S_{n}=3n^2-4n$, the sum of first n terms.
• What is the nth term?

 

If n is larger than 1, the sum of n terms includes anything below the first n-1 terms.

$a_{n}$ is the nth term of an A.P.

$\implies S_{n}-S_{n-1}=a_{n}\;(n\geq 2)$

 

The sums of terms:

• $S_{n}=3n^2-4n$

• $S_{n-1}=3n^2-10n+7$

$\implies a_{n}=6n-7\;(n\geq 2)$

 

The sum of the first 1 term only includes $a_{1}$.

$\implies S_{1}=a_{1}$

 

The first term:

$\implies a_{1}=-1$

 

So we can say $a_{n}=6n-7$.

 

More information:

Here is a known fact about $S_{n}$.

If $S_{n}$ doesn't contain a constant term, we can skip $S_1=a_{1}$ and say $S_{n}-S_{n-1}=a_{n}\;(n\geq 1)$.

If $S_{n}$ contains a constant term, $a_{1}$ is not actually the 1st term of A.P, so we must find it using $S_{1}=a_{1}$.

The second case is a series but not A.P because the difference changes.

For example, if $S_{n}=n^2+1$ the series is given as $a_{1}=2$, $a_{n}=2n-1\;(n\geq 2)$.

Answer 2

Question:

S_{n}=3n^2-4nS

n

=3n

2

−4n , the sum of first n terms.

What is the nth term?

 

If n is larger than 1, the sum of n terms includes anything below the first n-1 terms.

a_{n}a

n

is the nth term of an A.P.

\implies S_{n}-S_{n-1}=a_{n}\;(n\geq 2)⟹S

n

−S

n−1

=a

n

(n≥2)

 

The sums of terms:

S_{n}=3n^2-4nS

n

=3n

2

−4n

S_{n-1}=3n^2-10n+7S

n−1

=3n

2

−10n+7

\implies a_{n}=6n-7\;(n\geq 2)⟹a

n

=6n−7(n≥2)

 

The sum of the first 1 term only includes a_{1}a

1

.

\implies S_{1}=a_{1}⟹S

1

=a

1

 

The first term:

\implies a_{1}=-1⟹a

1

=−1

 

So we can say a_{n}=6n-7a

n

=6n−7 .

 

More information:

Here is a known fact about S_{n}S

n

.

If S_{n}S

n

doesn't contain a constant term, we can skip S_1=a_{1}S

1

=a

1

and say S_{n}-S_{n-1}=a_{n}\;(n\geq 1)S

n

−S

n−1

=a

n

(n≥1) .

If S_{n}S

n

contains a constant term, a_{1}a

1

is not actually the 1st term of A.P, so we must find it using S_{1}=a_{1}S

1

=a

1

.

The second case is a series but not A.P because the difference changes.

For example, if S_{n}=n^2+1S

n

=n

2

+1 the series is given as a_{1}=2a

1

=2 , a_{n}=2n-1\;(n\geq 2)a

n

=2n−1(n≥2) .