Question

2
Q.V.3) Find the 31" term of an arithmetic progression whose 11' term is 38 and the 16th
term is 73.​

Answer 1

Step-by-step explanation:

gave any other questions

Answer 2

Answer :

31st term = 178

Step-by-step explanation :

   $\underline{\underline{\bf Arithmetic \ Progression:}}$

• It is the sequence of numbers such that the difference between any two successive numbers is constant.

• In AP,

        a - first term

        d - common difference

        aₙ - nth term

        Sₙ - sum of n terms

• General form of AP,

           a , a+d , a+2d , a+3d , ..........

• nth term of AP,

         $\boxed{\bf a_n=a+(n-1)d}$

______________________________

Given,

11th term = 38

16th term = 73

To find,

31st term = ?

• 11th term = 38

       

       a + (11-1)d = 38

         a + 10d = 38 ----(1)

• 16th term = 73

       a + (16-1)d = 73

       a + 15d = 73 ----(2)

=> Substitute eq.(1) from eq.(2)

  a + 15d - (a + 10d) = 73 - 38

  a + 15d - a - 10d = 35

            5d = 35

              d = 35/5

              d = 7

common difference = 7

=> substitute the value of d in any eq.

     a + 10d = 38

     a + 10(7) = 38

     a + 70 = 38

     a = 38 - 70

     a = -32

first term = -32

=> 31st term :-

    $a_{31}=-32+(31-1)(7)\\a_{31}=-32+30(7) \\a_{31}=-32+210\\\boxed{a_{31}=178}$

31st term of AP = 178