Math, asked by mahanteshvarsha, 2 days ago

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

Answers

Answered by Anonymous
24

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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Answered by ItzFadedGuy
10

{\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.

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