How many terms of the ap: 9,17,25...must be taken to give a sum of 636

1.	How many terms of the ap: 9,17,25...must be taken to give a sum of 636
Answer» According to the question,we have,\xa0a=9. Therefore, common difference d =17-9=8let the required number of terms be n.Therefore, Sn=636{tex}\\Rightarrow{/tex}{tex}\\frac{n}{2}{/tex}[2a+(n-1)d]=636{tex}\\Rightarrow{/tex}{tex}\\frac{n}{2}{/tex}[2(9)+(n-1)8]=636{tex}\\Rightarrow{/tex}n[18+8n-8]=1272{tex}\\Rightarrow{/tex}8n2+10n-1272=0{tex}\\Rightarrow{/tex}4n2+5n-636=0{tex}\\Rightarrow{/tex}4n2+53n-48n-636=0{tex}\\Rightarrow{/tex}n(4n+53)-12(4n+53)=0{tex}\\Rightarrow{/tex}(4n+53)(n-12)=0{tex}\\Rightarrow{/tex}4n+53=0 or n-12=0{tex}\\Rightarrow{/tex}n={tex}\\frac{{ - 53}}{4}{/tex}\xa0or n=12Since number of terms cannot neither be negative nor fraction, n=12hence, the required number of terms is 12.

Discussion