Math, asked by YoWeGood, 3 months ago

What will be the second term of an AP if the sum of its first n terms is 2n2 +6n

Answers

Answered by user0888
2

Let S_n=2n^2+6n. And let the n-th term be a_n.

The relation between S_n and a_n is as follows.

  • S_1=a_1
  • S_n-S_{n-1}=a_n for n≥2

We want a_2, which is the second term.

Using the relation,

S_2-S_1=a_2

\Longleftrightarrow 20-8=a_2

\Longleftrightarrow 12=a_2

Hence, the answer is 12.

More information:

About S_n for two cases

  • S_n=an^2+bn (without a constant)

This satisfies

S_n-S_{n-1}=a_n over every natural numbers.

To prove this, you can simplify LHS. Then show they always equal by S_1=a_1.

  • S_n=an^2+bn+c (with a constant)

This satisfies

S_n-S_{n-1}=a_n for n≥2.

To prove this, you can simplify LHS. Then show they never equal for n=1 by S_1=a_1.

Answered by Anonymous
0

Answer:

Let S_n=2n^2+6nS

n

=2n

2

+6n . And let the n-th term be a_na

n

.

The relation between S_nS

n

and a_na

n

is as follows.

S_1=a_1S

1

=a

1

S_n-S_{n-1}=a_nS

n

−S

n−1

=a

n

for n≥2

We want a_2a

2

, which is the second term.

Using the relation,

S_2-S_1=a_2S

2

−S

1

=a

2

\Longleftrightarrow 20-8=a_2⟺20−8=a

2

\Longleftrightarrow 12=a_2⟺12=a

2

Hence, the answer is 12.

More information:

About S_nS

n

for two cases

S_n=an^2+bnS

n

=an

2

+bn (without a constant)

This satisfies

S_n-S_{n-1}=a_nS

n

−S

n−1

=a

n

over every natural numbers.

To prove this, you can simplify LHS. Then show they always equal by S_1=a_1S

1

=a

1

.

S_n=an^2+bn+cS

n

=an

2

+bn+c (with a constant)

This satisfies

S_n-S_{n-1}=a_nS

n

−S

n−1

=a

n

for n≥2.

To prove this, you can simplify LHS. Then show they never equal for n=1 by S_1=a_1S

1

=a

1

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