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In an AP, the pth term is q and ( p +q)...

In an AP, the pth term is q and ( p +q) term is 0. Then, prove that its qth term is p.

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Let a be the first term and d be the common difference of the given AP. Then ,
` T_(q) =q Rightarrow a+ ( p-1) d=q`
` and T_( p+q) =0 Rightarrow a+ ( p+q -1) d=0`
On subtracting (i) from (ii), we get
` qb = -q Rightarrow d=-1`
Putting d=-1 in (i) , we get a = ( q+p -1)
` T_(q) = a+ ( q-1) d= ( q+p-1)+ ( q -1) xx ( -1) Rightarrow T_(q) =p " " [ because a= ( q+p-1) and d=-1]`
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