Question

If $m$ times the ${m^{th}}$ term of an AP is equal to $n$ times its ${n^{th}}$ term, show that ${(m + n)^{th}}$ term is zero.

Answer 1

Hint- For solving this question, we should proceed by writing the ${m^{th}}$ term and ${n^{th}}$ and then we need to equate $m.({T_m}) = n.({T_n})$. By solving this expression, we will get the value of ${(m + n)^{th}}$ term.

Complete step-by-step answer:
According to the question, it is said that $m.({T_m}) = n.({T_n})$.
We know ${m^{th}}$ term is given by, ${T_m} = a + (m - 1)d$ and ${n^{th}}$ term is given by, ${T_n} = a + (n - 1)d$ .
Now it is said that $m$ times the ${m^{th}}$ term of an AP is equal to $n$ times its ${n^{th}}$ term.
$\Rightarrow m.\left[ {a + (m - 1)d} \right] = n.\left[ {a + (n - 1)d} \right]$
$\Rightarrow m.\left[ {a + (m - 1)d} \right] - n.\left[ {a + (n - 1)d} \right] = 0$
$\Rightarrow a(m - n) + d\left[ {(m + n)(m - n) - (m - n)} \right] = 0$
$\Rightarrow (m - n)\left[ {a + ((m + n) - 1)d} \right] = 0$ ….(1)
From equation (1), it can be said that,
$\Rightarrow a + ((m + n) - 1)d = 0$ …(2)
We know that
$\Rightarrow {T_{m + n}} = a + ((m + n) - 1)d$ ,
Hence, from equation (2), it can be said that ${T_{m + n}} = 0$.

Note- For solving questions based on terms of an AP, we need to use the general term of an AP as given, ${T_r} = a + (r - 1)d$ where $r$ denotes the number of terms in AP, $a$ denotes the initial term and $d$ denotes the common difference of the AP. Also, AP can be represented by the following series of terms; $a,a + d,a + 2d,a + 3d,...$