Determine k so that k+2, 4k-6 and 3k-2 are the three consecutive terms of an A.P.

Let the first term of the A.P. be $a$ and the common difference be $d$.
The $k^{th}$ term of an A.P. is given by the formula $a_k = a + (k-1)d$.
Therefore, the $m^{th}$ term is $a_m = a + (m-1)d$.
And the $n^{th}$ term is $a_n = a + (n-1)d$.

The problem states that $m$ times the $m^{th}$ term is equal to $n$ times the $n^{th}$ term:

Substitute the expressions for $a_m$ and $a_n$:

Group terms with $a$ and terms with $d$:

Factor out $a$ and $d$:

Rearrange the terms inside the bracket:

Factor the difference of squares $n^2 - m^2 = (n-m)(n+m)$:

Factor out $(n-m)$ from the right side:

Since $(n-m) = -(m-n)$, we can write:

Assuming $m \neq n$, we can divide both sides by $(m-n)$:

The $(m+n)^{th}$ term is $a_{m+n} = a + ((m+n)-1)d$.

Thus, the $(m+n)^{th}$ term of the A.P. is $\boxed{0}$.

2 हो गए। आज 3 बाकी — अगला भेजो।