In an arithmetic progression,if T_m = n and T | vedclass

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(A) Let the first term be $a$ and the common difference be $d$. The $n$-th term of an arithmetic progression is given by $T_n = a + (n - 1)d$. Given $

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$m + n - p$

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(A) Let the first term be $a$ and the common difference be $d$. The $n$-th term of an arithmetic progression is given by $T_n = a + (n - 1)d$. Given $T_m = a + (m - 1)d = n$  (Equation $1$) Given $T_n = a + (n - 1)d = m$  (Equation $2$) Subtracting Equation $2$ from Equation $1$: $(a + (m - 1)d) - (a + (n - 1)d) = n - m$ $(m - 1 - n + 1)d = n - m$ $(m - n)d = -(m - n)$ $d = -1$ Substituting $d = -1$ in Equation $1$: $a + (m - 1)(-1) = n$ $a - m + 1 = n$ $a = m + n - 1$ Now,find $T_p$: $T_p = a + (p - 1)d$ $T_p = (m + n - 1) + (p - 1)(-1)$ $T_p = m + n - 1 - p + 1$ $T_p = m + n - p$

In an arithmetic progression,if $T_m = n$ and $T_n = m$,then $T_p = \dots$

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