Subjective Type

If the sum of first $$p$$ terms of an A.P.is equal to the sum of the first $$q$$ terms, then find the sum of the first $$(p+q)$$ terms.

Solution

Let $$a$$ and $$d$$ be the first term and the common difference of the A.P. respectively.
Thus, $$\displaystyle { S }_{ p }=\frac { p }{ 2 } \left[

2a+(p-1)d \right]$$
and $${ S }_{ q }=\dfrac { q }{ 2 } \left[ 2a+(q-1)d

\right]$$
Now according to the given condition,
$$\dfrac

{ p }{ 2 } \left[ 2a+(p-1)d \right] =\dfrac { q }{ 2 } \left[ 2a+(q-1)d

\right] \quad \\ \Rightarrow p\left[ 2a+(p-1)d \right] =q\left[

2a+(q-1)d \right] \\ \Rightarrow 2ap+pd(p-1)=2aq+qd(q-1)\\ \Rightarrow

2a(p-q)+d\left[ p(p-1)-q(q-1) \right] =0\\ \Rightarrow 2a(p-q)+d\left[ {

p }^{ 2 }-p-{ q }^{ 2 }+q \right] =0\\ \Rightarrow 2a(p-q)+d\left[ {

(p-q)(p+q)-(p-q) } \right] =0\\ \Rightarrow 2a(p-q)+d\left[ {

(p-q)(p+q-1) } \right] =0\\ \Rightarrow 2a+d(p+q-1)=0\\ \Rightarrow

d=\dfrac { -2a }{ p+q-1 } \\ \therefore { S }_{ p+q }=\dfrac { p+q }{ 2 }

\left[ 2a+(p+q-1).d \right] \\ \quad =\dfrac { p+q }{ 2

} \left[ 2a+(p+q-1)\left( \dfrac { -2a }{ p+q-1 } \right) \right]=0$$

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