1.

The sum of p terms of an A.P. is q and the sum of q terms is p. Find the sum of terms (p+q)?

Answer» Let a be the first term and d the common difference of the given A.P.{tex}\\therefore S_{p}=\\frac{p}{2}{/tex}\xa0[2a + (p - 1)d] = q\xa0{tex}\\Rightarrow{/tex}\xa02a + (p - 1)d\xa0{tex}=\\frac{2 q}{p}{/tex} ….(i)And\xa0{tex}S_{q}=\\frac{q}{2}{/tex}\xa0[2a + (q - 1)d] = p{tex}\\Rightarrow{/tex}\xa02a + (q - 1)d\xa0{tex}=\\frac{2 p}{q}{/tex}\xa0….(ii)Subtracting eq. (ii) from eq. (i) we get(p - q)d =\xa0{tex}\\frac{2 q}{p}-\\frac{2 p}{q}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0(p - q)d\xa0{tex}=\\frac{2\\left(q^{2}-p^{2}\\right)}{p q}{/tex}{tex}\\Rightarrow{/tex}\xa0(p - q)d\xa0{tex}=\\frac{-2}{p q}{/tex}(p2\xa0- q2){tex}\\Rightarrow{/tex}\xa0(p - q)d\xa0{tex}=\\frac{-2}{p q}{/tex}\xa0(p + q)(p - q)\xa0{tex}\\Rightarrow d=\\frac{-2}{p q}{/tex}\xa0(p + q)Substituting the value of d in eq. (i) we get2a + (p - 1)\xa0{tex}\\left[\\frac{-2(p+q)}{p q}\\right]=\\frac{2 q}{p}{/tex}{tex}\\Rightarrow 2 a=\\frac{2 q}{p}+\\frac{2(p-1)(p+q)}{p q}{/tex}{tex}\\Rightarrow a=\\frac{q}{p}+\\frac{(p-1)(p+q)}{p q}{/tex}{tex}a=\\frac{q^{2}+p^{2}+p q-p-q}{p q}{/tex}Now\xa0Sp+q\xa0{tex}=\\frac{p+q}{2}{/tex}\xa0[2a + (p + q - 1)d{tex}=\\frac{p+q}{2}\\left[\\frac{2 q^{2}+2 p^{2}+2 p q-2 q-2 q}{p q}+\\frac{(p+q-1)[-2(p+q)}{p q}\\right]{/tex}{tex}=\\frac{p+q}{2}\\left[\\frac{2q^{2} + 2p^{2} + 2pq - 2p - 2q -2p^{2} -2 p q+2 p-2 p q-2 q^{2}+2 q}{p q}\\right]{/tex}{tex}=\\frac{p+q}{2}\\left[\\frac{-2 p q}{p q}\\right]{/tex}\xa0= -(p + q)\xa0hence proved.


Discussion

No Comment Found

Related InterviewSolutions