Question

Two APs have the same common difference. The difference between their 100th terms is 100,what is the difference between their 1000th terms?​

Answer 1

Given :-

• Two APs have the same common difference.
• The difference between their 100th terms is 100.

To Find :-

• what is the difference between their 1000th terms ?

Formula used :-

• nth Term of AP = a + ( n - 1)d

Solution :-

→ Lets First Term of First AP = a

→ Let Common Difference of First AP = d

So,

→ T₁₀₀ = a + (100 - 1)d

→ T₁₀₀ = a + 99d ------------ Equation (1)

→ Lets First Term of Second AP = b

→ Let Common Difference of Second AP = d

So,

→ T₁₀₀ = b + (100 - 1)d

→ T₁₀₀ = b + 99d ------------ Equation (2)

Given That, Equation (1) - Equation (2) = 100 .

So,

→ ( a + 99d) - (b + 99d) = 100

→ a - b + 99d - 99d = 100

→ (a - b) = 100 ------------------ Equation (3). .

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Now, we have to Find difference between their 1000th terms .

So,

→ T₁₀₀₀ = a + (1000 - 1)d ( First Series) .

→ T₁₀₀₀ = a + 999d ---------- Equation (4)

And,

→ T₁₀₀₀ = b + (1000 - 1)d (Second Series)

→ T₁₀₀₀ = b + 999d ----------------- Equation (5) .

So, Equation (4) - Equation (5) :-

→ (a + 999d) - (b + 999d)

→ (a - b) + (999d - 999d)

→ (a - b) + 0

→ (a - b)

Putting Value of Equation (3) Here , we get,

→ 100 (Ans.)

Answer 2

Given

A word problem from Ap

where

cd is same

To find

The difference between their 1000th terms

Solution

Let the first term of first ap be a and second be b

we know that

$\implies{T_n=a+(n-1)d}$

So,

$\implies{T_{100}}=a+(n-1)d}$

&

$\implies{T_{100}}=b+(n-1)d}$

According to the question

___________________________

$\implies{a+99d}$-(${b+99d}$)

$\implies{a+99d}$-(${b+99d}$)=100

=>a+99d-b-99d=100

=>a-b=100_____(1)

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Now for 1000 term

we have to do same steps

• Finding 1000 terms and subtracting it

________________________________

$\implies{a+999d}$-(${b+999d}$)

=>a+999d-b-999d

=>a-b

And from equation (1)

the value is 100

___________________________

Hence, The difference between their 1000th terms is 100