the fourth term of an AP is 5 while the sum of the first 6 terms is 10. Find the sum of the first 19 terms

S₁₉ = 855

Step-by-step explanation:

T₄ = a + ( n - 1 )d  = 5 , from the statement above , but n = 4

       a + 3d  = 5 1

S₆ = ⁿ/₂[(2a + ( n - 1 )d]  =  10, where n = 6

    = ⁶/₂( 2a + 5d )         = 10

    = 3( 2a + 5d ) = 10

    = 6a + 15d      = 10 2

Now solve the two equation together simultaneously to get the values of a and d

   a + 3d     = 5

   6a + 15d = 10

from 1,

a = 5 - 3d 3

Now put (3) in equation 2 and open the brackets

6( 5 - 3d )  + 15d = 10

30 - 18d + 15d      = 10

30 - 3d                 = 10

            3d            = 30 - 10

             3d           = 20

                         d = ²⁰/₃.

Now substitute for d to get a in equation 3

           a = 5 - 3( ²⁰/₃)

           a = 5 - 3 ₓ ²⁰/₃

              = 5 - 20

          a  = -15.

Now to find the sum of the first 19 terms,

we use the formula

S₁₉ = ⁿ/₂( 2a + ( n - 1 )d )

     = ¹⁹/₂( 2 x -15 + 18 x ²⁰/₃ )

     = ¹⁹/₂( -30 + 6 x 20 )

     = ¹⁹/₂( -30 + 120 )

     = ¹⁹/₂( 90 )

     = ¹⁹/₂ x 90

     = 19 x 45

     = 855

Therefore,

S₁₉ = 855

the answer is 62.8 cubic units

the formula of a slope:

we have the points (-2, 7) and (2, 3). substitute: