Question

The sum of first and 21st terms of an arithmetic sequence is 140.
a) Find the sum of 6th term and 16th term.
b) what is the 11th term.
c) Find the sum of first 21 terms.
d) Find the sum of first 11 terms of the sequence 20, 25, 30.​

Answer 1

Given :

The sum of first and 21st terms of an arithmetic sequence is 140.

To Find :

• The sum of 6th term and 16th term.
• The 11th term.
• The sum of first 21 terms.
• The sum of first 11 terms of the sequence 20, 25, 30.

Solution :

We know that for general term of an AP the formula is,

aₙ = a + (n - 1)d

where,

• a = First term
• n = Term no
• d = Common Difference

ATQ,

• First term = a ------ (i)

21st term :

⇒ aₙ = a + (n - 1)d

⇒ a₂₁ = a + (21 - 1)d

⇒ a₂₁ = a + (20)d

⇒ a₂₁ = a + 20d

∴ a₂₁ = a + 20d ------ (ii)

Adding (i) & (ii),

⇒ a + a₂₁ = 140

⇒ a + a + 20d = 140

⇒ 2a + 20d = 140

Dividing both sides by 2,

⇒ a + 10d = 70

∴ a + 10d = 70.

(i) The sum of 6th term and 16th term :

6th term,

⇒ aₙ = a + (n - 1)d

⇒ a₆ = a + (6 - 1)d

⇒ a₆ = a + (5)d

⇒ a₆ = a + 5d

∴ a₆ = a + 5d. ----- (iii)

16th term,

⇒ aₙ = a + (n - 1)d

⇒ a₁₆ = a + (16 - 1)d

⇒ a₁₆ = a + (15)d

⇒ a₁₆ = a + 15d

∴ a₁₆ = a + 15d. ----- (iv)

Adding (iii) & (iv),

= a₆ + a₁₆

= a + 5d + a + 15d

= 2a + 20d

Now, from the above calculation we can see that

= 2a + 20d = 140.

So,

Sum of 6th term and 16th term is 140.

(ii) The 11th term :

From above calculation we can see that,

⇒ a + 10d = 70

We know that,

⇒ a₁₁ = a + 10d

⇒ a₁₁ = 70

11th term is 70.

(iii) The sum of first 21 terms :

We know that,

Sₙ = n/2[2a + (n - 1)d]

where,

• a = First term
• n = Term no
• d = Common Difference

⇒ S₂₁ = 21/2[2a + (21 - 1)d]

⇒ S₂₁ = 21/2[2a + (20)d]

⇒ S₂₁ = 21/2[2a + 20d]

From above calculation we know that (2a + 20 = 140),

⇒ S₂₁ = 21/2[140]

⇒ S₂₁ = 21/2 × 140

⇒ S₂₁ = 21 × 70

⇒ S₂₁ = 1470

∴ S₂₁ = 1470.

Sum of first 21 terms is 1470.

(iv) The sum of first 11 terms of the sequence 20, 25, 30 :

• a = 20
• d = a₂ - a₁ = 25 - 20 = 5
• n = 11

So,

Sₙ = n/2[2a + (n - 1)d]

⇒ S₁₁ = 11/2[2(20) + (11 - 1)5]

⇒ S₁₁ = 11/2[2 × 20 + (10)5]

⇒ S₁₁ = 11/2[40 + 10 × 5]

⇒ S₁₁ = 11/2[40 + 50]

⇒ S₁₁ = 11/2[90]

⇒ S₁₁ = 11/2 × 90

⇒ S₁₁ = 11 × 45

⇒ S₁₁ = 495

∴ S₁₁ = 495.

Sum of first 11 terms of sequence 20, 25, 30..., is 495.

Answer 2

Step-by-step explanation:

Given :

The sum of first and 21st terms of an arithmetic sequence is 140.

To Find :

The sum of 6th term and 16th term.

The 11th term.

The sum of first 21 terms.

The sum of first 11 terms of the sequence 20, 25, 30.

Solution :

We know that for general term of an AP the formula is,

aₙ = a + (n - 1)d

where,

a = First term

n = Term no

d = Common Difference

ATQ,

First term = a ------ (i)

21st term :

⇒ aₙ = a + (n - 1)d

⇒ a₂₁ = a + (21 - 1)d

⇒ a₂₁ = a + (20)d

⇒ a₂₁ = a + 20d

∴ a₂₁ = a + 20d ------ (ii)

Adding (i) & (ii),

⇒ a + a₂₁ = 140

⇒ a + a + 20d = 140

⇒ 2a + 20d = 140

Dividing both sides by 2,

⇒ a + 10d = 70

∴ a + 10d = 70.

(i) The sum of 6th term and 16th term :

6th term,

⇒ aₙ = a + (n - 1)d

⇒ a₆ = a + (6 - 1)d

⇒ a₆ = a + (5)d

⇒ a₆ = a + 5d

∴ a₆ = a + 5d. ----- (iii)

16th term,

⇒ aₙ = a + (n - 1)d

⇒ a₁₆ = a + (16 - 1)d

⇒ a₁₆ = a + (15)d

⇒ a₁₆ = a + 15d

∴ a₁₆ = a + 15d. ----- (iv)

Adding (iii) & (iv),

= a₆ + a₁₆

= a + 5d + a + 15d

= 2a + 20d

Now, from the above calculation we can see that

= 2a + 20d = 140.

So,

Sum of 6th term and 16th term is 140.

(ii) The 11th term :

From above calculation we can see that,

⇒ a + 10d = 70

We know that,

⇒ a₁₁ = a + 10d

⇒ a₁₁ = 70

11th term is 70.

(iii) The sum of first 21 terms :

We know that,

Sₙ = n/2[2a + (n - 1)d]

where,

a = First term

n = Term no

d = Common Difference

⇒ S₂₁ = 21/2[2a + (21 - 1)d]

⇒ S₂₁ = 21/2[2a + (20)d]

⇒ S₂₁ = 21/2[2a + 20d]

From above calculation we know that (2a + 20 = 140),

⇒ S₂₁ = 21/2[140]

⇒ S₂₁ = 21/2 × 140

⇒ S₂₁ = 21 × 70

⇒ S₂₁ = 1470

∴ S₂₁ = 1470.

Sum of first 21 terms is 1470.

(iv) The sum of first 11 terms of the sequence 20, 25, 30 :

a = 20

d = a₂ - a₁ = 25 - 20 = 5

n = 11

So,

Sₙ = n/2[2a + (n - 1)d]

⇒ S₁₁ = 11/2[2(20) + (11 - 1)5]

⇒ S₁₁ = 11/2[2 × 20 + (10)5]

⇒ S₁₁ = 11/2[40 + 10 × 5]

⇒ S₁₁ = 11/2[40 + 50]

⇒ S₁₁ = 11/2[90]

⇒ S₁₁ = 11/2 × 90

⇒ S₁₁ = 11 × 45

⇒ S₁₁ = 495

∴ S₁₁ = 495.

Sum of first 11 terms of sequence 20, 25, 30..., is 495.