Answer: 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
