Question

The sum of the first 9 terms of an arithmetic sequence is 270. What is the 5th term of this sequence? Write the sequence if it's common difference is 7.

Answer 1

Answer:

30

2, 9, 16, 23, ...

Step-by-step explanation:

Using Sₙ = (n/2) [2a + (n - 1)d]

   In the given question,

⇒ S₉ = (9/2) [2a + (9 - 1)d]

⇒ 270 = (9/2) [2a + 8d]

⇒ 270 = (9/2) 2(a + 4d)

⇒  270 = 9 (a + 4d)

⇒ 270/9 = a + 4d

⇒ 30 = a + 4d

        Note that 5th term = a + (5 - 1)d

                                        = a +  4d

                     ∴ 5th term = 30

If the common difference is 7,

⇒ 5th term = 30   ⇒ a + (5 - 1)d = 30

⇒ a + 4d = 30

⇒ a + 4(7) = 30        [d = 7]

⇒ a = 2

    Hence the sequence is:

a,   a + d,   a + 2d,    a + 3d, ...

2,   2 + 7,   2 + 2(7),  2 + 3(7), ...

2, 9, 16, 23, ...

Answer 2

★ Given:

• The sum of the first 9 terms of an arithmetic sequence is 270.

• Common difference is 7.

★ To find:

• The 5th term of the sequence.

• We have to write the sequence.

★ Solution:

᯾ Finding 5th term of the sequence

We know that,

$\sf{\pink{\underbrace{\underline{S_n = \Big( \frac{n}{2} \Big) \Big[2a + (n-1)d \Big]}}}}$

$\sf{S_9 = \Big( \frac{9}{2} \Big) \Big[2a + (9 - 1)d \Big] }$

$\sf{270 = \Big( \frac{9}{2} \Big) \Big[2a + 8d \Big] }$

$\sf{270 = \Big( \frac{9}{2}\Big)2(a + 4d)}$

$\sf{270 =9(a + 4d)}$

$\sf{ \frac{270}{9} = a + 4d}$

$\sf{30 = a + 4d}$

᯾ Finding sequence

$\sf{a + 4d = 30}$

$\sf{a + 4(7) = 30}$

$\sf{= 2}$

• Therefore, the sequence is

$\sf{a, \: a + d, \: a + 2d, \: a + 3d \:---}$

$\sf{2, \: 2+7, 2+2(7), \: 2+3(7) \:---}$

$\sf{2, \: 9, \: 16, \: ,23 \:---}$