Answer
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Hint: Following the question we will get two equations. Subtracting them we will get the value of d and putting d value in one equation we will get the first term of the sequence. Substituting them in the general formula we will get the answer.
Complete step-by-step answer:
We know that in arithmetic progression, the general formula is,
${a_n} = a + (n - 1)d$
Where, ${a_n}$ = ${n^{th}}$ term of AP
a = first term of the AP
d = common difference in AP
Thus ${m^{th}}$ term, ${a_m} = a + (m - 1)d$
In the question it is given that ${m^{th}}$ term is n
i.e. $a + (m - 1)d = n$………………….(1)
And ${n^{th}}$ term, ${a_n} = a + (n - 1)d$
It is also given in the question that ${n^{th}}$ term is m
I.e. $a + (n - 1)d = m$………………..(2)
Subtracting equation (2) from equation (1) we find the common difference of the arithmetic series.
Hence, $[a + (m - 1)d] - [a + (n - 1)d] = n - m$
$ \Rightarrow a + (m - 1)d - a - (n - 1)d = n - m$
Cancelling a and –a in left hand side we get,
$(m - 1)d - (n - 1)d = n - m$
Taking d common in left hand side we get,
$(m - 1 - n + 1)d = n - m$
Cancelling -1 and +1 we get,
$(m - n)d = n - m$
$ \Rightarrow d = \dfrac{{n - m}}{{m - n}}$
Multiplying -1 on both side we get,
$d = - 1$
Putting value of d in equation (2) we get,
$a + (n - 1)d = m$
$ \Rightarrow a + (n - 1)\left( { - 1} \right) = m$
$ \Rightarrow a - n + 1 = m$
$ \Rightarrow a = m + n - 1$
We got the value of a and d.
For the ${p^{th}}$ term,
We will use the general formula of AP i.e.
${a_n} = a + (n - 1)d$
Putting n = p, a = m+n-1 and d = -1 in the above formula we get,
${a_p} = \left( {m + n - 1} \right) + \left( {p - 1} \right) - 1$
Expanding the right hand side of the equation we get,
${a_p} = m + n - 1 - p + 1$
Cancelling -1 and +1 in the right hand side we get,
${a_p} = m + n - p$
Thus the ${p^{th}}$ term is ${a_p} = m + n - p$.
Note: Arithmetic progression or arithmetic sequence is the sequence in which the difference between two consecutive numbers are equal.
You can also subtract equation 1 from equation 2 to get the d value.
Be cautious while doing the equations because the mistakes in minus and plus signs can even change the whole answer.
Complete step-by-step answer:
We know that in arithmetic progression, the general formula is,
${a_n} = a + (n - 1)d$
Where, ${a_n}$ = ${n^{th}}$ term of AP
a = first term of the AP
d = common difference in AP
Thus ${m^{th}}$ term, ${a_m} = a + (m - 1)d$
In the question it is given that ${m^{th}}$ term is n
i.e. $a + (m - 1)d = n$………………….(1)
And ${n^{th}}$ term, ${a_n} = a + (n - 1)d$
It is also given in the question that ${n^{th}}$ term is m
I.e. $a + (n - 1)d = m$………………..(2)
Subtracting equation (2) from equation (1) we find the common difference of the arithmetic series.
Hence, $[a + (m - 1)d] - [a + (n - 1)d] = n - m$
$ \Rightarrow a + (m - 1)d - a - (n - 1)d = n - m$
Cancelling a and –a in left hand side we get,
$(m - 1)d - (n - 1)d = n - m$
Taking d common in left hand side we get,
$(m - 1 - n + 1)d = n - m$
Cancelling -1 and +1 we get,
$(m - n)d = n - m$
$ \Rightarrow d = \dfrac{{n - m}}{{m - n}}$
Multiplying -1 on both side we get,
$d = - 1$
Putting value of d in equation (2) we get,
$a + (n - 1)d = m$
$ \Rightarrow a + (n - 1)\left( { - 1} \right) = m$
$ \Rightarrow a - n + 1 = m$
$ \Rightarrow a = m + n - 1$
We got the value of a and d.
For the ${p^{th}}$ term,
We will use the general formula of AP i.e.
${a_n} = a + (n - 1)d$
Putting n = p, a = m+n-1 and d = -1 in the above formula we get,
${a_p} = \left( {m + n - 1} \right) + \left( {p - 1} \right) - 1$
Expanding the right hand side of the equation we get,
${a_p} = m + n - 1 - p + 1$
Cancelling -1 and +1 in the right hand side we get,
${a_p} = m + n - p$
Thus the ${p^{th}}$ term is ${a_p} = m + n - p$.
Note: Arithmetic progression or arithmetic sequence is the sequence in which the difference between two consecutive numbers are equal.
You can also subtract equation 1 from equation 2 to get the d value.
Be cautious while doing the equations because the mistakes in minus and plus signs can even change the whole answer.
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