• # question_answer 3) How many terms of an A.P. 9, 17, 25, .., must be taken to give a sum of 636?

 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)$ $n=12$