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8. Sequences and Series
easy
If the ${p^{th}}$ term of an $A.P.$ be $\frac{1}{q}$ and ${q^{th}}$ term be $\frac{1}{p}$, then the sum of its $p{q^{th}}$ terms will be
A
$\frac{{pq - 1}}{2}$
B
$\frac{{1 - pq}}{2}$
C
$\frac{{pq + 1}}{2}$
D
$ - \frac{{pq + 1}}{2}$
Solution
(c) Since ${T_p} = a + (p – 1)d = \frac{1}{q}$…..$(i)$
and ${T_q} = a + (q – 1)d = \frac{1}{p}$ …..$(ii)$
From $(i)$ and $(ii),$ we get $a = \frac{1}{{pq}}$ and $d = \frac{1}{{pq}}$
Now sum of $pq$ terms $ = \frac{{pq}}{2}\left[ {\frac{2}{{pq}} + (pq – 1)\frac{1}{{pq}}} \right]$
$ = \frac{{pq}}{2}.\frac{2}{{pq}}\left[ {1 + \frac{1}{2}(pq – 1)} \right] = \left[ {\frac{{2 + pq – 1}}{2}} \right] = \frac{{pq + 1}}{2}$
Note : Students should remember this question as a formula.
Standard 11
Mathematics
