question_answer

How many terms of an A.P. 9, 17, 25, .., must be taken to give a sum of 636?

Answer:

A.P. is \[9,17,25,.....,{{S}_{n}}=636\]

\[a=9,d=17-9=8\]

\[{{S}_{n}}=\frac{n}{2}[2a+(n-1)d]\]

\[636=\frac{n}{2}[18+8n-8]\]

\[636=\frac{n}{2}[10+8n]\]

\[636=n(5+4n)\]

\[636=5n+4{{n}^{2}}\]

\[4{{n}^{2}}+5n-636=0\]

\[4{{n}^{2}}+53n-48n-636=0\]

\[n(4n+53)-12(4n+53)=0\]

\[(n-12)(4n+53)=0\]

\[n-12=0\]\[\left( \because \,n\ne \frac{-53}{4}as\,\,n>0 \right)\]

\[n=12\]