Question

the sum of the fourth and eight term of an A.P is 22 and the product of the second and the sixth is 33 find A.P​

Answer 1

GIVEN :-

• Sum of the $\sf 4^{th}$ and $\sf 8^{th}$ term  of an AP = 22

• Product of  $\sf 2^{nd}$ and $\sf 6^{th}$ term = 33

TO FIND :-

• AP

SOLUTION :-

 Let ,

  $\sf 1^{st}$ term = a

  Common difference = d

           $\bf \boxed{\bf a_n=a+(n-1)d}$

   

  $\bullet \sf \ a_4+a_8 = 22$

     $\longrightarrow \sf a+(4-1)d +a+(8-1)d = 22 \\\\\longrightarrow a+3d+a+7d = 22 \\\\\longrightarrow 2a+10d= 22 \\\\\longrightarrow a+5d= 11 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..............(1)$

 $\bullet\sf \ a_2\times a_6 = 33$

   $\longrightarrow\sf a+(2-1)d \times a+(6-1)d =33 \\\\\longrightarrow (a+d)\times (a+5d) = 33$

   We know that a + 5d = 11 ,

   $\longrightarrow\sf (a+d)\times 11 = 33 \\\\\longrightarrow a+d = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..............(2)$

 Equation (1) - Equation (2) ,

  $\longrightarrow \sf 4d = 8 \\\\\longrightarrow \bf d = 2$

  Substitute d = 2 in equation (2) ,

   $\longrightarrow\sf a+2= 3 \\\\\longrightarrow \bf a = 1$

   AP = 1 , 3 , 5 , 7 , 9 , 11 , .....................