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If the sum of p terms of an AP is q and the sum of q terms is p, then show that the sum of `p+q` terms is `- (p+q)` , Also find the sum of first `p-q` terms (where , `pgtq`). |
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Answer» let first term common difference of the AP be and d respectively . then `S_(p)=q` `implies (P)/(2)[2a+(p-1)d]=q` ` 2a+(p-1)d=(2q)/(p)` and `S_(q)-p` `implies (q)/(2)[2a+(q-1)d]=p` `implies 2a+(q-1)d=(2p)/(q)` On subtracting Eq (ii) from Eq (i) we get `2a+(p-1)d-2a-(q-1)d=(2q)/(p)-(2p)/(q)` ` implies [(p-1)-(q-1)]d=(2p^(2)-2p^(2))/(pq)` `implies [p-1-q_1]d=(2(q^(2)-p^(2)))/(pq)` `implies (p-q)d=(2(q^(2)-p^(2)))/(pq)` `therefore d=(-2(p+q))/(pq)` On substiuting the value of d in (eq (i) we get ` 2a+(p-1)((-2(p+q))/(pq))=(2q)/(p)` `implies 2a=(2q)/(p)+(2(P+q)(P-1))/(pq)` `implies a=[(2q)/(p) +(2(p+q)(p-1))/(pq)]` `Now S_(p+q) -(p+q)/(2)[2a+(p+q-1)d]` `=(p+q)/(2)[(2q)/(p)+(2(p+q)(p-1))/(pq)-((p+q-1)2(P+q))/(pq)]` ` =(p+q)[(q)/(p)+((p+q(p-1)-(p+q-1)(P+q))/(pq)]` `=(p+q)[(q)/(p)+((p+q)(p-1)-(p+q-1)(P+q))/(pq)]` `=(P+q)[(q)/(p)+((p+q)(p-1-p-q+1)(p+q))/pq)]` `=p+q [(q)/(p)-(p+q)/(p)]=(p+q)[(q-p-q)/(p)]` `S_(p+q)=-(p+q)` `S_(p-q)=(p-q)/(2)[2a+(p-q-1)d]` `=(p-q)/(2)[(2p)/(p)+(2(p+q)(p-1))/(pq)-(p-q-1)2(p+q))/(pq)]` `=(p-q)[(q)/(p) +(p+q(p-1-p+q+1))/(pq)]` `=(p-q)[(q)/(p)+((p+q)q)/(pq)]` `=(p-q)[(q)/(p)+(p+q)/(p)]=(p-q)((p+2q))/(p)` |
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