Question

Is 5,8,11,14,.........an A.P.?If so then what will be the 100th term?check whether 92 is in A.P. ?Is no. 61 is in A.P.​

Answer 1

Answer:

1. The given sequence is an AP.

2. The 100ᵗʰ term of the AP is 302.

3. 92 is the 30ᵗʰ term of the AP.

4. The number 61 is not in the AP.

Step-by-step-explanation:

The given sequence is 5, 8, 11, 14,...

Here,

• t₁ = 5
• t₂ = 8
• t₃ = 11
• t₄ = 14

Now,

→ t₂ - t₁ = 8 - 5 = 3

→ t₃ - t₂ = 11 - 8 = 3

→ t₄ - t₃ = 14 - 11 = 3

Here, the difference between two consecutive terms is constant i. e. 3.

∴ The given sequence is an AP.

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Now,

We know that,

• a = 5
• d = 3

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

→ t₁₀₀ = 5 + ( 100 - 1 ) * 3

→ t₁₀₀ = 5 + 99 * 3

→ t₁₀₀ = 5 + 297

→ t₁₀₀ = 302

∴ The 100ᵗʰ term of the AP is 302.

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Now,

We have to check whether 92 is in the AP or not.

Let the nᵗʰ term of the AP be 92.

Using the formula, tₙ = a + ( n - 1 ) d , we get,

→ 92 = 5 + ( n - 1 ) * 3

→ 92 - 5 = ( n - 1 ) * 3

→ 87 = ( n - 1 ) * 3

→ n - 1 = 87 ÷ 3

→ n - 1 = 29

→ n = 29 + 1

→ n = 30

∴ 92 is the 30ᵗʰ term of the AP.

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Now,

We have to find whether 61 is in the AP or not.

Let the nᵗʰ term of the AP be 61.

Using the formula, tₙ = a + ( n - 1 ) d , we get,

→ 61 = 5 + ( n - 1 ) * 3

→ 61 - 5 = ( n - 1 ) * 3

→ 56 = ( n - 1 ) * 3

→ n - 1 = 56 ÷ 3

→ n = ( 56 ÷ 3 ) + 1

As the value of n is a fraction, we can say that,

∴ The number 61 is not in the AP.

Answer 2

⭐Given:

5,8,11,14.................(Given Sequence)

⭐To Find:

✨ Is 5,8,11,14,..........an A. P.?

✨ t100=?

✨ 92 is in A. P.?

✨61 is in A. P.?

⭐Solution:

(1) 5,8,11,14,...................(Given Sequence)

Here,

t1=a=5

t2=8

t3=11

t4=14

So, Common Difference (d)

→ t2-t1=8-5=3

→t3-t2=11-8=3

→t4-t3=14-11=3

Here, Common difference(d) is 3 which is constant . ............... (1)

Hence, 5,8,11,14 is an A. P.

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(2) tn=a+(n-1) d............. (Formula)

t100=5+(100-1) 3

→ t100=5+(99) 3..

→ t100= 5+297

→ t100→ 302

Hence, Hundredth term is 302.

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(3) Let, tn=92

tn=a+(n-1) d.......... (formula)

→92=5+(n-1) 3

→92=5+3n-3

→92=5-3+3n

→92=2+3n

→92-2=3n

→90=3n

→n=90/3

→n=30

30 th term of a given A. P is 92. Hence, 92 is in given A. P

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(4) Let tn=61

tn=a+(n-1) d............ (Formula)

→61=5+(n-1) 3

→61=5+3n-3

→ 61=5-3+3n

→ 61=2+3n

→ 61-2=3n

→59=3n

→ n=59/3

→n=19.6

But terms are not in decimal or point.

Hence 61 is not in an Given A. P..

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