Appropriate Question :-
If the pth term of an AP is 1/q and the qth term is 1/p, show that the sum of the pq terms is 1/2(pq+1)
Step-by-step explanation:
Let assume that first term and common difference of an AP is a and d respectively.
Given that,
[tex]\qquad\sf \: a_p = \dfrac{1}{q} \\ \\ [/tex]
[tex]\sf\implies \sf \: a + (p - 1)d = \dfrac{1}{q} - - - (1) \\ \\ [/tex]
Further given that,
[tex]\qquad\sf \: a_q = \dfrac{1}{p} \\ \\ [/tex]
[tex]\sf\implies \sf \: a + (q - 1)d = \dfrac{1}{p} - - - (2) \\ \\ [/tex]
On Subtracting equation (2) from equation (1), we get
[tex]\qquad\sf \: (p - 1)d - (q - 1)d = \dfrac{1}{q} - \dfrac{1}{p} \\ \\ [/tex]
[tex]\qquad\sf \: (p -1 - q + 1)d = \dfrac{p - q}{qp} \\ \\ [/tex]
[tex]\qquad\sf \: (p - q)d = \dfrac{p - q}{pq} \\ \\ [/tex]
[tex]\sf\implies \sf \: d = \dfrac{1}{pq} - - - (3) \\ \\ [/tex]
On substituting the value of d in equation (1), we get
[tex]\qquad\sf \: a + (p - 1) \times \dfrac{1}{pq} = \dfrac{1}{q} \\ \\ [/tex]
[tex]\qquad\sf \: a + \dfrac{1}{q} - \dfrac{1}{pq} = \dfrac{1}{q} \\ \\ [/tex]
[tex]\qquad\sf \: a - \dfrac{1}{pq} = 0 \\ \\ [/tex]
[tex]\sf\implies \sf \: a = \dfrac{1}{pq} - - - (4) \\ \\ [/tex]
Now, Consider
[tex]\sf \: S_{pq} \\ \\ [/tex]
[tex]\qquad\sf \: = \: \dfrac{pq}{2} \bigg(2a + (pq - 1)d \bigg) \\ \\ [/tex]
[tex]\qquad\sf \: = \: \dfrac{pq}{2} \bigg(2 \times \dfrac{1}{pq} + (pq - 1) \times \dfrac{1}{pq} \bigg) \\ \\ [/tex]
[tex]\qquad\sf \: = \: \dfrac{pq}{2} \times \dfrac{1}{pq} \bigg(2 + (pq - 1) \bigg) \\ \\ [/tex]
[tex]\qquad\sf \: = \: \dfrac{1}{2} \bigg(2 + pq - 1\bigg) \\ \\ [/tex]
[tex]\qquad\sf \: = \: \dfrac{1}{2} \bigg( pq + 1\bigg) \\ \\ [/tex]
Hence,
[tex]\sf\implies \bf \: S_{pq} \: = \: \dfrac{1}{2} \bigg( pq + 1\bigg) \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Formulae Used
↝ nᵗʰ term of an arithmetic progression is,
[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}[/tex]
↝ Sum of n terms of an arithmetic progression is,
[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}[/tex]
Wʜᴇʀᴇ,
a is the first term of the progression.
n is the no. of terms.
d is the common difference.
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Verified answer
Appropriate Question :-
If the pth term of an AP is 1/q and the qth term is 1/p, show that the sum of the pq terms is 1/2(pq+1)
Step-by-step explanation:
Let assume that first term and common difference of an AP is a and d respectively.
Given that,
[tex]\qquad\sf \: a_p = \dfrac{1}{q} \\ \\ [/tex]
[tex]\sf\implies \sf \: a + (p - 1)d = \dfrac{1}{q} - - - (1) \\ \\ [/tex]
Further given that,
[tex]\qquad\sf \: a_q = \dfrac{1}{p} \\ \\ [/tex]
[tex]\sf\implies \sf \: a + (q - 1)d = \dfrac{1}{p} - - - (2) \\ \\ [/tex]
On Subtracting equation (2) from equation (1), we get
[tex]\qquad\sf \: (p - 1)d - (q - 1)d = \dfrac{1}{q} - \dfrac{1}{p} \\ \\ [/tex]
[tex]\qquad\sf \: (p -1 - q + 1)d = \dfrac{p - q}{qp} \\ \\ [/tex]
[tex]\qquad\sf \: (p - q)d = \dfrac{p - q}{pq} \\ \\ [/tex]
[tex]\sf\implies \sf \: d = \dfrac{1}{pq} - - - (3) \\ \\ [/tex]
On substituting the value of d in equation (1), we get
[tex]\qquad\sf \: a + (p - 1) \times \dfrac{1}{pq} = \dfrac{1}{q} \\ \\ [/tex]
[tex]\qquad\sf \: a + \dfrac{1}{q} - \dfrac{1}{pq} = \dfrac{1}{q} \\ \\ [/tex]
[tex]\qquad\sf \: a - \dfrac{1}{pq} = 0 \\ \\ [/tex]
[tex]\sf\implies \sf \: a = \dfrac{1}{pq} - - - (4) \\ \\ [/tex]
Now, Consider
[tex]\sf \: S_{pq} \\ \\ [/tex]
[tex]\qquad\sf \: = \: \dfrac{pq}{2} \bigg(2a + (pq - 1)d \bigg) \\ \\ [/tex]
[tex]\qquad\sf \: = \: \dfrac{pq}{2} \bigg(2 \times \dfrac{1}{pq} + (pq - 1) \times \dfrac{1}{pq} \bigg) \\ \\ [/tex]
[tex]\qquad\sf \: = \: \dfrac{pq}{2} \times \dfrac{1}{pq} \bigg(2 + (pq - 1) \bigg) \\ \\ [/tex]
[tex]\qquad\sf \: = \: \dfrac{1}{2} \bigg(2 + pq - 1\bigg) \\ \\ [/tex]
[tex]\qquad\sf \: = \: \dfrac{1}{2} \bigg( pq + 1\bigg) \\ \\ [/tex]
Hence,
[tex]\sf\implies \bf \: S_{pq} \: = \: \dfrac{1}{2} \bigg( pq + 1\bigg) \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Formulae Used
↝ nᵗʰ term of an arithmetic progression is,
[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}[/tex]
↝ Sum of n terms of an arithmetic progression is,
[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}[/tex]
Wʜᴇʀᴇ,
a is the first term of the progression.
n is the no. of terms.
d is the common difference.