Question

In an A.P., if pth term is 1/q and qth term is 1/p , prove that the sum of first pq terms is 1/2(pq+1), where p is not equal to q..
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Answer 1

$\large\bf\underline{Given:-}$

• pth term = 1/q
• qth term = 1/p

$\large\bf\underline {To \: prove:-}$

sum of first pq terms is 1/2(pq+1)

$\huge\bf\underline{Solution:-}$

$\blacktriangleright \bf \large \: a_n = a + (n - 1)d$

• pth term is 1/q

$\rightarrow \rm \: a_p = a + (p - 1)d \\ \\ \rightarrow \rm \: a + (p - 1)d = \frac{1}{q}........(i)$

• qth term is 1/p

$\rightarrow \rm \:a_q = a + (q - 1)d \\ \\ \rightarrow \rm \: a+(q-1)d=\frac{1}{p}......(ii)$

• Substracting eq.(ii) from (i)

$\rm a + (p - 1)d = \frac{1}{q}\\ \rm \: a+(q-1)d = \frac{1}{p} \\ - \: \: \: \: \: \: \: \: \: \: \: \: \: \: - \: \: \: \: \: \: \: \: \: \: - \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \rm \: \: \leadsto \: (p - 1)d - (q - 1)d = \frac{1}{q} - \frac{1}{p} \\ \rm \: \: \: \: \: \: \: \leadsto \: pd - d - qd + d = \frac{p - q}{pq} \\ \: \: \: \: \: \: \: \: \rm \leadsto \: pd - qd = \frac{p - q}{pq} \\ \: \: \: \: \: \: \: \: \: \rm \leadsto \: (p - q)d = \frac{p - q}{pq} \\ \: \: \: \: \: \: \: \: \: \rm \leadsto \: d = \frac{ \cancel{(p - q)}}{pq} \times \frac{1}{ \cancel{(p - q)}} \\ \: \: \: \: \: \: \: \: \bf \leadsto \: d = \frac{1}{pq}$

Putting value of d = 1/pq in eq.(i)

$\rm :\implies \: a + (p - 1)d = \frac{1}{q} \\ \\ \rm :\implies \: a + (p - 1) \times \frac{1}{pq} = \frac{1}{q} \\ \\ \rm :\implies \: a + \frac{p}{pq} - \frac{1}{pq} = \frac{1}{q} \\ \\ \rm :\implies \: a + \frac{1}{q} - \frac{1}{pq} = \frac{1}{q} \\ \\ \rm :\implies \: a = \cancel \frac{1}{q} - \cancel\frac{1}{q} + \frac{1}{pq} \\ \\ \bf:\implies \: a = \frac{1}{pq}$

• So, a = 1/pq and d = 1/pq

$\large \blacktriangleright \bf \: S_n= \frac{n}{2} [2a + (n - 1)d]$

Sum of 1st pq terms :-

$\rm \dashrightarrow \: S_{pq} = \frac{pq}{2} [2 \times \frac{1}{pq} + (pq - 1) \times \frac{1}{pq} ] \\ \\ \rm \dashrightarrow \: S_{pq} = \frac{pq}{2} [ \frac{2}{pq} + \frac{pq}{pq} - \frac{1}{pq} ] \\ \\ \rm \dashrightarrow \: S_{pq} = \frac{pq}{2} [ \frac{2 + pq- 1}{pq} ] \\ \\ \rm \dashrightarrow \: S_{pq} = \frac{pq}{2} [ \frac{1 + pq}{pq} ] \\ \\ \rm \dashrightarrow \: S_{pq} = \frac{ \cancel{pq}}{2} \{ \frac{1 + pq}{ {\cancel{pq}}} \} \\ \\ \bf \dashrightarrow \: S_{pq} = \frac{1}{2} \{ pq + 1\}$

Hence proved that the sum of first pq terms is 1/2(pq+1), where p is not equal to q.

Answer 2

$\sf\orange{Given:}$

$\sf{\implies{t_{p}=\frac{1}{q}}}$

$\sf{\implies{t_{q}=\frac{1}{p}}}$

$\sf{\implies{p \ is \ not \ equal \ to \ q}}$

$\sf\pink{To \ prove:}$

$\sf{S_{pq}=\frac{1}{2}(pq+1)}$

$\sf\green{\underline{\underline{Proof:}}}$

$\boxed{\sf{t_{n}=a+(n-1)d}}$

$\sf{\therefore{t_{p}=a+(p-1)d}}$

$\sf{\therefore{\frac{1}{q}=a+(p-1)d}}$

$\sf{\therefore{a+(p-1)d=\frac{1}{q}...(1)}}$

$\sf{t_{q}=a+(q-1)d}$

$\sf{\therefore{\frac{1}{p}=a+(q-1)d}}$

$\sf{\therefore{a+(q-1)d=\frac{1}{p}...(2)}}$

$\sf{Subtract \ equation (1) \ from \ equation (2)}$

$\sf{a+(q-1)d=\frac{1}{p}}$

$\sf{-}$

$\sf{a+(p-1)d=\frac{1}{q}}$

_________________________

$\sf{(q-1)d-(p-1)d=\frac{1}{p}-\frac{1}{q}}$

$\sf{\therefore{d(q-1-p+1)=\frac{q-p}{qp}}}$

$\sf{\therefore{d(q-p)=\frac{p-q}{qp}}}$

$\sf{\therefore{d=\frac{p-q}{qp}\times\frac{1}{p-q}}}$

$\sf{\therefore{d=\frac{1}{qp}}}$

$\boxed{\sf{\therefore{d=\frac{1}{pq}...(3)}}}$

$\sf{Substitute \ d=\frac{1}{pq} \ in \ equation (1)}$

$\sf{a+(p-1)\times\frac{1}{pq}=\frac{1}{q}}$

$\sf{\therefore{a+\frac{(p-1)}{pq}=\frac{1}{q}}}$

$\sf{\therefore{a=\frac{1}{q}-\frac{(p-1)}{pq}}}$

$\sf{\therefore{a=\frac{p-(p-1)}{pq}}}$

$\sf{\therefore{a=\frac{p-p+1}{pq}}}$

$\boxed{\sf{\therefore{a=\frac{1}{pq}...(4)}}}$

_________________________________

$\boxed{\sf{S_{n}=\frac{n}{2}[2a+(n-1)d]}}$

$\sf{\therefore{S_{pq}=\frac{pq}{2}[2a+(pq-1)d]}}$

$\sf{But, \ a=\frac{1}{pq} \ and \ d=\frac{1}{pq}}$

$\sf{....from \ (3) \ and \ (4)}$

$\sf{\therefore{S_{pq}=\frac{pq}{2}[\frac{2}{pq}+(pq-1)\times\frac{1}{pq}]}}$

$\sf{\therefore{S_{pq}=\frac{pq}{2}[\frac{2}{pq}+\frac{pq-1}{pq}]}}$

$\sf{\therefore{S_{pq}=\frac{pq}{2}[\frac{2+(pq-1)}{pq}]}}$

$\sf{\therefore{S_{pq}=\frac{1}{2}(2+pq-1)}}$

$\sf{\therefore{S_{pq}=\frac{1}{2}(pq+1)}}$

$\sf{Hence, \ proved.}$

$\sf\purple{\tt{Sum \ of \ first \ pq^{th} \ term \ is \ \frac{1}{2}(pq+1).}}$