Question

the pth term of an AP is q and qth term is p.find it's ( p+q ) th term​

Answer 1

Solution -

Firstly,

➝ Let the first term of A.P. be a.

➝ Let the common difference of A.P. be d.

⠀

It is given that,

➝ a + (p - 1)d = q⠀⠀⠀...[1]

➝ a + (q - 1)d = p⠀⠀⠀...[2]

⠀

Subtracting [1] from [2]

➝ [a + (q - 1)d] - [a + (p - 1)d] = (p - q)

➝ [a + dq - d] - [a + dp - d] = (p - q)

➝ a + dq - d - a - dp + d = (p - q)

➝ dq - dp = (p - q)

➝ d(q - p) = (p - q)

➝ d = -1

⠀

Putting the value of d in [1]

➝ a + (p - 1)(-1) = q

➝ a - p + 1 = q

➝ a = (p + q - 1)

⠀

Now, we have

• First term = (p + q -1)
• Common difference = -1

⠀

Finding the (p + q)th term

➝ a + (p + q - 1)d

➝ (p + q - 1) + (p + q - 1)(-1)

➝ (p + q - 1) - (p + q - 1)

➝ 0

⠀

Hence,

• (p + q)th term of the given is 0.

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Answer 2

${\large{\bold{\sf{\underline{Given \; that:}}}}}$

• pth term of A.P is q
• qth term of A.P is p

${\large{\bold{\sf{\underline{To \; find:}}}}}$

• (p+q) th term

${\large{\bold{\sf{\underline{Solution:}}}}}$

To find the nth term, we generally use the formula:

$\tt{:\implies a_n = a+(n-1)d}$

We are going to apply this formula for pth term and qth term as well.

$\tt{:\implies a_p = q}$

$\tt{:\implies a+(p-1)d = q}$

$\tt{:\implies a+pd-d = q\: ---(1)}$

We are assuming this as Equation (1). We also know that:

$\tt{:\implies a_q = p}$

$\tt{:\implies a+(q-1)d = p}$

$\tt{:\implies a+qd-d = p\: ---(2)}$

We are assuming this as Equation (2).

On subtracting Equation (2) from Equation (1), we get:

$\tt{:\implies (a+pd-d) - (a+qd-d) = q-p}$

$\tt{:\implies a+pd-d -a-qd+d = q-p}$

$\tt{:\implies pd-qd = q-p}$

$\tt{:\implies d(p-q) = q-p}$

$\tt{:\implies d = \dfrac{q-p}{p-q}}$

$\tt{:\implies -d = \dfrac{p-q}{p-q}}$

$\tt{:\implies -d = 1}$

$\tt{:\implies d = -1}$

Hence, the value of d is 1-. Let us substitute our value in Equation (1).

$\tt{:\implies a+pd-d = q}$

$\tt{:\implies a+p(-1)-(-1) = q}$

$\tt{:\implies a-p+1 = q}$

$\tt{:\implies a = p+q-1}$

We know that,

$\tt{:\implies a_{p+q} = a+(p+q-1)d}$

$\tt{:\implies a_{p+q} = a+(p+q-1)d}$

$\tt{:\implies a_{p+q} = a+ad}$

$\tt{:\implies a_{p+q} = a+a \times -1}$

$\tt{:\implies a_{p+q} = a-a}$

$\tt{:\implies a_{p+q} = 0}$

${\large{\bold{\sf{\underline{Conclusion:}}}}}$

Hence, (p+q) th term of an A.P is 0.