The third term of an AP is 8 and ninth term AP exceeds three times the third term by 2. Find the sum of its nineteenth terms

a+2d = 8

a+8d = 3*8+2

so, a=2 and d=3

Now you want

S19 = 19/2 (2*2 + 18*3)

Not satisfied.

T(3)=8 - - - - - - - (1)

& T(9)=3*T(3)+2 - - - - - - - - - (2)
Find S(19)

T(n) = a+(n-1) d
Hence equation (1) becomes,
a+(3-1)d=8
a+2d=8
a=8-2d - - - - - - - - - - (3)

Also equation (2) becomes,
a+(9-1) d = 3×[a+(3-1) d] +2
a+8d=[3*(a+2d)]+2
a+8d=(3a+6d)+2
a+8d=3a+6d+2
Rearranging,
8d-6d=3a-a+2
2d=2a+2
Dividing by 2 on both sides,
d=a+1 - - - - - - - - - - - - - - - - - - - (4)
Substituting (3) in (4) we get,
d=(8-2d)+1
d=8-2d+1
d=9-2d
3d=9
d=9/3
d=3

Substituting value of d in (3)
a=8-2(3)
a=8-6
a=2

Hence a=2 & d=3

We know that, S(n) = n/2 × [2a+(n-1) d]
S(19) = 19/2 × [2*2+(19-1)*3]
= 19/2 * [4+18*3]
= 19/2 * [4+54]
=19/2 * 58
=19*29
=551

Sum of first 19 terms of grid AP is 551

To find the sum of the nineteenth terms of the arithmetic progression (AP), we first need to find the common difference (d) and the first term (a).

Given that the third term (a3) is 8:
a3 = a + 2d = 8 (Equation 1)

Also, the ninth term (a9) exceeds three times the third term (a3) by 2:
a9 = 3a3 + 2 = 3(8) + 2 = 26 (Equation 2)

To solve these equations simultaneously, we can substitute Equation 1 into Equation 2:
a + 2d = 8 (Equation 1)
3(a + 2d) + 2 = 26 (Equation 2)

Simplifying Equation 2, we get:
3a + 6d + 2 = 26
3a + 6d = 24 (Equation 3)

Now, let's solve equations 1 and 3 simultaneously:
a + 2d = 8 (Equation 1)
3a + 6d = 24 (Equation 3)

To solve this system of equations, we can use the method of substitution or elimination. Let's use elimination.

Multiply Equation 1 by 3 and subtract Equation 3 from it:
(3a + 6d) - (3a + 6d) = 24 - 24
0 = 0

Since the equations result in 0 = 0, this means that the system is dependent or consistent, meaning there are infinitely many solutions. We cannot determine unique values for a and d.

However, we can still find the sum of the arithmetic progression's nineteenth terms (S19) by using the formula:

S19 = (n/2) * (2a + (n-1)d)

Here, n represents the number of terms. Since we want to find the sum of the 19 terms (S19), we can substitute n = 19 into the formula:

S19 = (19/2) * (2a + (19-1)d)

S19 = 9.5 * (2a + 18d)

Since we don't have unique values for a and d, we cannot calculate the exact sum of the nineteenth terms of the arithmetic progression without further information.