Question

The mean of 5th and 11th term of an AP is 25 and sum of 3rd and 9th term is 40.
Then find the 20th term.

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Answer 1

Let 'a' be the first term and 'd' be the common difference of the AP

Using nth term of AP formula

aₙ = a + ( n - 1 )d

5th term of AP = a + 4d

11th term of AP = a + 10d

Given :

Mean of 5th term and 11th term of an AP = 25

⇒  ( a + 4d + a + 10d ) / 2 = 25

⇒ ( 2a + 14d ) / 2 = 25

⇒ [ 2( a + 7d ) ] / 2 = 25

⇒ a + 7d = 25 → ( 1 )

Using nth term of AP formula

aₙ = a + ( n - 1 )d

3rd term = a + 2d

9th term = a + 8d

Given :

Sum of 3rd term and 9th term = 40

⇒ a + 2d + a + 8d = 40

⇒ 2a + 10d = 40

⇒ 2( a + 5d ) = 40

⇒ a + 5d = 40/2

⇒ a + 5d = 20 →  ( 2 )

Subtracting ( 2 ) from  ( 1 ) we get

• a + 7d = 25
• a + 5d = 20
• (-)  (-)     (-)

________________

• 2d = 5

________________

⇒ d = 5/2

Substituting d = 5/2 in ( 1 ) we get

⇒ a + 7d = 25

⇒ a + 7( 5/2 )  = 25

⇒ a + 35/2 = 25

⇒ a = 25 - 35/2

⇒ a = ( 50 - 35 ) / 2

⇒ a = 15/2

Using nth term of AP formula

aₙ = a + ( n - 1 )d

20th term = a₂₀ = a + 19d

⇒ a₂₀ = 15/2 + 19(5/2 )

⇒ a₂₀ = 15/2 + 95/2

⇒ a₂₀ = 110/2 = 55

Therefore the 20th term of the AP is 55.

Answer 2

Step-by-step explanation:

_______________________________

$\bf \underline{Question}$

The mean of 5th and 11th term of an AP is 25 and sum of 3rd and 9th term is 40.

Then find the 20th term.

_______________________________

$\bf \underline{Given}$

• The mean of 5th and 11th term of an AP is 25
• sum of 3rd and 9th term is 40.
• 5th term = a + 4d

_______________________________

$\bf \underline{To\:Find}$

• find the 20th term.

_______________________________

Applying in the formula

aₙ = a + ( n - 1 )d

5th term of AP

aₙ = a+( 5 - 1)d

aₙ = a+ 4d

5th term of AP = a + 4d

11th term of AP

aₙ = a+( 11 - 1)d

aₙ = a+ 10 d

11th term of AP = a + 10d

According to the Question:-

⇒  ( a + 4d + a + 10d ) / 2 = 25

⇒ ( 2a + 14d ) / 2 = 25

⇒ [ 2( a + 7d ) ] / 2 = 25

⇒ a + 7d = 25 ....equ. 1.

Applying formula

aₙ = a + ( n - 1 )d

3rd term

aₙ = a + ( 3 - 1 )d

aₙ = a + 2d

3rd term = a + 2d

9th term

aₙ = a + ( 9 - 1 )d

aₙ = a + 8d

9th term = a + 8d

According to the Question:-

⇒ a + 2d + a + 8d = 40

On adding

⇒ 2a + 10d = 40

⇒ 2( a + 5d ) = 40

On Transferring

⇒ a + 5d = 40/2

⇒ a + 5d = 20 .......equ.2

Subtracting equ. ( 2 ) from  equ.( 1 ) we get

a + 7d = 25

a + 5d = 20

________________

2d = 5

d = 5/2

Applying d value in equ. 1

⇒ a + 7d = 25

putting all values

⇒ a + 7( 5/2 )  = 25

⇒ a + 35/2 = 25

⇒ a = 25 - 35/2

⇒ a = ( 50 - 35 ) / 2

on substraction

⇒ a = 15/2

Applying in the formula

aₙ = a + ( n - 1 )d

20th term

aₙ = a + ( 20 - 1 )d

aₙ = a + 19d

20th term = a₂₀ = a + 19d

⇒ a₂₀ = 15/2 + 19(5/2 )

⇒ a₂₀ = 15/2 + 95/2

⇒ a₂₀ = 110/2 = 55

Answer is 55