Question

$\huge\blue{\boxed{Question}}$

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

Answer ASAP :)​

Answer 1

Answer:

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Step-by-step explanation:

here

Given that 2 ap's have same common difference

given that their 100th terms difference is 100

let the first no. of first series be a1 and second series be a2

then, a(1)100 - a(2)100=100 ---- 1

for 1st series ---- a100=a1+99d

   2nd series ---- a100 = a2+99d

keep these values in (1)

then,  

a1+99d - (a2+99d) = 100

a1+99d-a2-99d=100

therefore, a1-a2 =100 ------------------------------------------- 2

then the difference between their 1000th terms is

for 1st series --- a1000 = a1+999d

for 2nd series --- a1000 = a2+999d

their 100th terms difference is

a(1)1000-a(2)1000

a1+999d-(a2+999d)

a1+999d-a2-999d

therefore we get the value a1-a2

from (2) a1-a2 = 100

therefore the difference between their 1000th terms is 100

Answer 2

$\huge\blue{\boxed{Question}}$

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

$\huge\blue{\boxed{Answer....}}$

Let the 2 AP's be A1 and A2

It's given that A1 (100) - A2 (100) = 100 -------- 1

WE KNOW THAT A1(100) = A1 + 99D

A2(100) = A2 + 99D

=> SUBSTITUTING IN EQUATION 1 , WE GET ,

A1+99D-A2-99D = 100 '

SO WE GET A1-A2 = 100 --------------2

NOW WHEN WE TAKE TGE NEXT PART OF THE QUESTION , IT SAYS THAT

A1(1000) - A2(1000) = ?----------3

A1+999D= A1(1000)

A2+999D=A2(1000)

SUBSTITUTING IN EQUATION 3 , WE GET

A1+999D-A2-999D

=> A1-A2

AS WE KNOW THAT A1-A2 =100 ( EQUATION 2)

THEREFORE THE DIFFERENCE IS ALSO 100

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