Question

the sum of first 20 terms of an arithmetic sequence is 1060 it's 5 th term is 20 find the 15 th term
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Answer 1

Answer:

The 15ᵗʰ term of AP is 80.

Step-by-step-explanation:

We have given that,

The sum of first 20 terms of an A.P. is 1060.

The 5ᵗʰ term of AP is 20.

Now, we know that,

Sₙ = n / 2 [ 2a + ( n - 1 ) * d ] - - [ Formula ]

⇒ S₂₀ = 20 / 2 [ 2a + ( 20 - 1 ) * d ]

⇒ 1060 = 10 ( 2a + 19d )

⇒ 2a + 19d = 1060 ÷ 10

⇒ 2a + 19d = 106 - - ( 1 )

Now, we know that,

tₙ = a + ( n - 1 ) * d - - [ Formula ]

⇒ t₅ = a + ( 5 - 1 ) * d

⇒ 20 = a + 4d

⇒ a + 4d = 20 - - ( 2 )

By multiplying equation ( 2 ) by 2, we get,

⇒ 2 × ( a + 4d ) = 20 × 2

⇒ 2a + 8d = 40 - - ( 3 )

By subtracting equation ( 3 ) from equation ( 1 ), we get,

⇒ 2a + 19d - 2a - 8d = 106 - 40

⇒ 11d = 66

⇒ d = 66 ÷ 11

⇒ d = 6

By substituting d = 6 in equation ( 2 ), we get,

⇒ a + 4d = 20 - - ( 2 )

⇒ a + 4 ( 6 ) = 20

⇒ a + 24 = 20

⇒ a = 20 - 24

⇒ a = - 4

Now, by using the formula tₙ = a + ( n - 1 ) * d,

⇒t₁₅ = a + ( 15 - 1 ) * d

⇒ t₁₅ = ( - 4 ) + 14 × 6

⇒ t₁₅ = - 4 + 84

⇒ t₁₅ = 80

∴ The 15ᵗʰ term of AP is 80.

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Additional Information:

Arithmetic Progression:

1. In a sequence, if the common difference between two consecutive terms is constant, then the sequence is called as Arithmetic Progression ( AP ).

2. nᵗʰ term of an AP:

The number of a term in the given AP is called as ] term of an AP.

3. Formula for nᵗʰ term of an AP:

• tₙ = a + ( n - 1 ) * d

4. The sum of the first n terms of an AP:

The addition of either all the terms of a particular terms is called as sum of first n terms of AP.

5. Formula for sum of the first n terms of A. P. :

• Sₙ = n / 2 [ 2a + ( n - 1 ) * d ]

Answer 2

⭐GIVEN:

• Sum of first 20 terms of an arithmetic sequence is 1060. I. e S20=1060

• fifth term(t5) =20

⭐TO FIND:

• Fifteenth term (t15) =?

⭐SOLUTION:

Since we know that,

Sn=n/2[2a+(n-1) d]....... (formula for finding sum of terms)

: . S20=20/2[2a+(20-1) d]

: . 1060=10[2a+19d]

: . 1060/10=2a+19d

: . 2a+19d=106...................(1)

tn=a+(n-1) d..... (formula for finding the n th term)

: . t5=a+(5-1) d

: . 20=a+4d

: . a+4d=20.................. (2)

Multiplying equation (2) by 2 we get,

2a+8d=40.................(3)

Subtracting eq. (3) from (1) we get,

11d=66

: . d=66/11

: . d=6

Substituting the value of d in (2) we get,

a+4d=20............. (2)

: . a+4(6) =20

: . a+24=20

: . a=20-24

: . a=-4

Now we have the first term a=-4 and the common difference d=6

So let's find the 15 th term.

According to the formula,

t15=-4+(15-1) 6

: . t15= -4+(14×6)

: . t15=-4+ 84

: . t15= 80

Therefore the 15 the term of an Arithmetic sequence is 80.