Question

the APS have the same common difference if difference between their 10th term is hundred then the difference between their fourth term is​

Answer 1

Given: The Arithmetic progressions have same common difference. The Difference between their 10th term is 100.

Need to Find: The difference between their fourth term.

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We know,

$\: \: \: \: \: \: \: \: \: \tt{{a}_{n}\: =\: a\: +\: (n-1)\: d}$

Where,

$\sf{{a}_{n}\: is\: general\: term}$

$\sf{a\: is\: First\: term}$

$\sf{n\: is\: term\: number}$

$\sf{d\: is\: common\: difference}$

❍ Let's first find the 10th term of both the APs.

$\\ \underline{\large{\textsf{First AP :}}}$

⟩ So, 10th term will be,

• Let the First term of First AP be 'a'

$\: \: \: \: \: \: \: \longrightarrow\sf{{a}_{10}=a+(10-1)d}$

$\: \: \: \: \: \: \: \longrightarrow\sf{{a}_{10}=a+9d}$

$\bullet$ Therefore, 10th term of First AP = a+9d

$\\ \underline{\large{\textsf{Second AP :}}}$

• Let the First term of Second AP be 'b'

$\: \: \: \: \: \: \: \implies\sf{{b}_{n}=b+(n-1)d}$

$\: \: \: \: \: \: \: \implies\sf{{b}_{10}=b+(10-1)d}$

$\: \: \: \: \: \: \: \implies\sf{{b}_{10}=b+9d}$

$\bullet$ Therefore, 10th term of Second AP= b+9d

We are given that,

$\: \: \textsf{Difference between their 10th term is 100}$

$\: \: \: \: \: \: \: \dashrightarrow\sf{{a}_{10}-{b}_{10}=100}$

$\: \: \: \: \: \: \: \dashrightarrow\sf{(a+9d)-(b+9d)=100}$

$\: \: \: \: \: \: \: \dashrightarrow\sf{a+9d-b+9d=100}$

$\: \: \: \: \: \: \: \dashrightarrow\sf{a-b+9d-9d=100}$

$\: \: \: \: \: \: \: \dashrightarrow\sf{a-b=100\: \: \: -(1)}$

Now, we need to find the difference between their fourth term.

$\: \: \: \: \: \: \: \twoheadrightarrow\sf{{a}_{4}-{b}_{4}}$

$\: \: \: \: \: \: \: \twoheadrightarrow\sf{[a+(4-1)d]-[b+(4-1)d]}$

$\: \: \: \: \: \: \: \twoheadrightarrow\sf{[a+3d]-[b+3d]}$

$\: \: \: \: \: \: \: \twoheadrightarrow\sf{a+3d-b-3d}$

$\: \: \: \: \: \: \: \twoheadrightarrow\sf{a-b}$

From (1),

$\: \: \: \: \: \: \: \twoheadrightarrow\sf{a-b=100}$

So, difference between 4th term is 100.