If the {p^{th}} term of an A.P. be frac{1}{q} | vedclass

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Question

(c) Since ${T_p} = a + (p - 1)d = \frac{1}{q}$&hellip;..$(i)$

and ${T_q} = a + (q - 1)d = \frac{1}{p}$ &hellip;..$(ii)$

From $(i)$ and $(ii),$ w

Vedclass · Accepted Answer

$\frac{{pq + 1}}{2}$

Vedclass · Answer

(c) Since ${T_p} = a + (p - 1)d = \frac{1}{q}$&hellip;..$(i)$

and ${T_q} = a + (q - 1)d = \frac{1}{p}$ &hellip;..$(ii)$

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.

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

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