In AP, if s6+s7=167 And s10 =235 then find the AP ,where sn denote the sum of its first n term

In AP, if s6+s7=167 And s10 =235 then find the AP ,where sn denote the

1.	In AP, if s6+s7=167 And s10 =235 then find the AP ,where sn denote the sum of its first n term
Answer» Sn =\xa0{tex}\\frac{n}{2}{/tex}[2a+(n-1)d]Now, S5+S7=167{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{5}{2}{/tex}[2a+4d]+{tex}\\frac{7}{2}{/tex}[2a+6d]=167{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{{5 \\times 2}}{2}{/tex}[a+2d]+\|{tex}\\frac{{7 \\times 2}}{2}{/tex}[a+3d]=167{tex}\\Rightarrow{/tex}\xa05a+10d+7a+21d=167{tex}\\Rightarrow{/tex}\xa012a+31d=167...(i)Also, S10=235{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{{10}}{2}{/tex}[2a+9d]=235{tex}\\Rightarrow{/tex}\xa05[2a+9d]=235{tex}\\Rightarrow{/tex}\xa010a+45d=235{tex}\\Rightarrow{/tex}\xa02a+9d=47...(ii)Multiplying equation (ii) by 6, we get12a+54d=282...(iii)subtracting (i) from (iii), we get\xa023d = 115{tex}\\Rightarrow{/tex}\xa0d = 5{tex}\\Rightarrow{/tex}\xa02a+9(5)=47{tex}\\Rightarrow{/tex}\xa02a = 2{tex}\\Rightarrow{/tex}\xa0a = 1First term = a = 1Second term = a+d=1+5=6Third term = a+2d=1+2(5)=11Thus, the Ap is 1,6,11,...