Question

19. In A.P sum of the first 10
term is 155 and sum of the first
20 term is 610 then find the
sum of 11th to 20th term. *​

Answer 1

Step-by-step explanation:

we get s20= 610

so our ans is correct

Answer 2

Answer: Sum of 11th to 20th term is 423

Step-by-step explanation

a = ?, d= ?,  $S_{10} = 155$,  $S_{20} = 610$

$S_{n} = \frac{n}{2}*(2a + (n-1)d)$

$S_{10}$ = 10/2 (2a + (10-1)d)

155 = 10/2 (2a+9d)

310 = 20a +90d -- (1)

$S_{n} = \frac{n}{2}*(2a + (n-1)d)$

$S_{20}$ = 20/2 (2a + (20-1)d)

610 = 20/2 (2a + 19d)

1220 = 40a + 380d --(2)

          20a + 90d = 310 --(1) x2

         40a + 380d = 1220 --(2) x1

        ------------------------------

         40a + 180d = 620

         40a + 380d = 1220

         -----------------------------

         -       -               -

         -----------------------------

                  -200d = -600

                         d  =  3

20a + 90d = 310

20a   + 90x3 = 310

20a + 270 = 310

     20a      = 310 - 270

      20a      =     40

         a         =    2

To find sum of 11th to 20th term

$S_{n} = \frac{n}{2}*(2a + (n-1)d)$

$S_{11}$ = 11/2 ( 2a + (11-1)d)

$S_{11}$ =  11/2 (2x2 + 10x3)

$S_{11}$ =    11/2 ( 4 + 30)

$S_{11}$ =   11/2 x 34

$S_{11}$ =   187

Sum of 11th to 20th term =   $S_{20} - S_{11}$

                                           = 610 - 187

                                           =  423

Hope this helps!!

Cheers!!