Question

The 10th and 15th terms of an AP are 37
and 47 respectively. Find the
sum of first
20 terms of the series.​

Answer 1

GIVEN :-

• 10th term of an AP = 37

• 15th term of an AP  = 47

TO FIND :-

• Sum of first 20 terms of the AP .

SOLUTION :-

    Let first term be and the common difference be d .

               $\bf a_n=a+(n-1)d$

• 10th term = 37

         $\longrightarrow\sf a+(10-1)d=37\\\\\longrightarrow a+9d = 37 \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..........(1)$

• 15th term = 47

        $\longrightarrow\sf a+(15-1)d=47\\\\\longrightarrow a+14d=47 \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..........(2)$

   Eq (2) - Eq (1) ,

       $\longrightarrow\sf 5d= 10\\\\\longrightarrow \bf d= 2$

 

  Substitute the value of d in Eq (1) ,

      $\longrightarrow\sf a+9\times2=37 \\\\\longrightarrow a+18=37\\\\\longrightarrow a=37-18\\\\\longrightarrow \bf a= 19$

  $\bf Sum \ of \ first \ n \ terms = \dfrac{n}{2}\times (2a+(n-1)d)$

    Sum of first 20 terms ,

     

            $\longrightarrow \sf \dfrac{20}{2}\times (2\times 19+(20-1)\times 2)\\\\\longrightarrow 10\times (38+19\times 2)\\\\\longrightarrow 10 \times (38+38)\\\\\longrightarrow 10\times 76 \\\\\longrightarrow\bf 760$