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| 1. |
In anAP if nth term is 1/n and mth term is 1/n find sum of first mn terms. |
| Answer» Let the first term = a and common difference = dGiven , am=\xa0{tex}\\frac{1}{n}{/tex}\xa0and an ={tex}\\frac{1}{m}{/tex}So,\xa0{tex}{\\mathrm a}_{\\mathrm m}=\\mathrm a+(\\mathrm m-1)\\mathrm d=\\frac1n {/tex}.........(1)and,{tex}{\\mathrm a}_{\\mathrm n}=\\mathrm a+(\\mathrm n-1)\\mathrm d=\\frac1m {/tex}.........(2)Subtracting (2) from (1), we get :-(m-n)d =\xa0{tex}\\frac{m-n}{mn}{/tex}Or, d={tex}\\frac{1}{mn}{/tex}Now putting value of d in (2),\xa0{tex}\\mathrm a+(\\mathrm n-1)\\times\\frac1{\\mathrm{mn}}=\\frac1{\\mathrm m}{/tex}{tex}\\mathrm a+\\frac1{\\mathrm m}-\\frac1{\\mathrm{mn}}=\\frac1{\\mathrm m}{/tex}{tex}\\mathrm a=\\frac1{\\mathrm{mn}}{/tex}Now,\xa0{tex}{\\mathrm S}_{\\mathrm{mn}}=\\frac{\\mathrm{mn}} {2}\\lbrack\\frac2{\\mathrm{mn}}+(\\mathrm{mn}-1)(\\frac1{\\mathrm{mn}})\\rbrack{/tex}{tex}\\Rightarrow {/tex}\xa0Smn\xa0=\xa0{tex}\\frac {1} {2}{/tex} [ 2 + (mn - 1)] 【taking (1/mn) common 】{tex}\\Rightarrow{/tex}\xa0Smn\xa0=\xa0{tex}\\frac 1 2{/tex}\xa0[1 + mn ]\xa0 | |