Question

if m times the mth term of an AP is equal to n times its nth term. prove that the ( m+n)th term of the AP is zero.​

Answer 1

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Answer 2

Answer:

We have to prove that $\sf\:t_{m\:+\:n}\:=\:0$

Step-by-step-explanation:

Let first term of AP be a and common difference be d.

From given condition,

$\sf\:m\:\times\:t_{m}\:=\:n\:\times\:t_{n}$

But, we know that,

$\sf\:t_{n}\:=\:a\:+\:(\:n\:-\:1\:)\:d\:\:-\:-\:[\:Formula\:]$

Similarly,

$\sf\:t_{m}\:=\:a\:+\:(\:m\:-\:1\:)\:d\\\\\implies\sf\:m\:\times\:[\:a\:+\:(\:m\:-\:1\:)\:d\:]\:=\:n\:\times\:[\:a\:+\:(\:n\:-\:1\:)\:d\\\\\implies\sf\:ma\:+\:md\:(\:m\:-\:1\:)\:=\:na\:+\:nd\:(\:n\:-\:1\:)\\\\\implies\sf\:ma\:+\:m^{2}d\:-\:md\:=\:na\:+\:n^{2}d\:-\:nd\\\\\implies\sf\:ma\:-\:na\:+\:m^{2}d\:-\:n^{2}d\:-\:md\:+\:nd\:=\:0\\\\\implies\sf\:a\:(\:m\:-\:n\:)\:+\:d\:(\:m^{2}\:-\:n^{2}\:)\:-\:d\:(\:m\:-\:n\:)\:=\:0\:\:\:-\:-\:[\:\because\:a^{2}\:-\:b^{2}\:=\:(\:a\:+\:b\:)\:(\:a\:-\:b\:)\:]\\\\\implies\sf\:(\:m\:-\:n\:)\:[\:a\:+\:d\:(\:m\:+\:n\:)\:-\:d\:]\:=\:0\\\\\implies\sf\:(\:m\:-\:n\:)[\:a\:+\:d\:(\:m\:+\:n\:)\:-\:d\:\times\:1\:]\:=\:0\\\\\implies\sf\:a\:+\:[\:(\:m\:+\:n\:)\:-\:1\:]\:d\:=\:0\:\:\:-\:-\:(\:1\:)$

Now,

$\sf\:t_{m\:+\:n}\:,\:here\:n^{th}\:term\:=\:m\:+\:n\\\\\therefore\sf\:t_{n}\:=\:a\:+\:(\:n\:-\:1\:)\:d\:\:-\:-\:[\:Formula\:]\\\\\implies\sf\:t_{m\:+\:n}\:=\:a\:+\:(\:m\:+\:n\:-\:1\:)\:d\\\\\implies\sf\:t_{m\:+\:n}\:=\:0\:\:\:-\:-\:-\:[\:From\:(\:1\:)\:]$

Additional Information:

1. 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. $\sf\:n^{th}$ term of an AP:

The number of a term in the given AP is called as $\sf\:n^{th}$ term of an AP.

3. Formula for $\sf\:n^{th}$ term of an AP:

$\sf\:t_{n}\:=\:a\:+\:(\:n\:-\:1\:)\:d$

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. :

$\boxed{\sf\:S_{n}=\frac{n}{2} [ 2a + ( n - 1) d ]}$