The third term of an AP is 8 and the ninth term of the AP exceeds three times the third term by 2 find the sum of its first 19 terms
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 first 19 terms of an arithmetic progression (AP), we need to know the first term (a) and the common difference (d).
Given:
- The third term (a + 2d) = 8
- The ninth term (a + 8d) = 3(a + 2d) + 2
Let's solve this system of equations to find the values of a and d.
1. The third term: a + 2d = 8
2. The ninth term: a + 8d = 3(a + 2d) + 2
Expanding the second equation:
a + 8d = 3a + 6d + 2
Rearranging the equation:
2a + 2d = 8d + 2
Simplifying:
2a - 6d = 2
Now, we can solve these simultaneous equations:
From equation (1): a = 8 - 2d
Substituting into equation (2):
2(8 - 2d) - 6d = 2
16 - 4d - 6d = 2
-10d = -14
d = (-14) / (-10)
d = 7/5
Substituting the value of d back in equation (1):
a = 8 - 2(7/5)
a = 40/5 - 14/5
a = 26/5
So, the first term (a) is 26/5 and the common difference (d) is 7/5.
Now, we can use the formula to find the sum of the first 19 terms of an AP:
Sum(n) = (n/2)(2a + (n - 1)d)
Substituting the values, where n = 19:
Sum(19) = (19/2)(2((26/5)) + (19 - 1)(7/5))
Sum(19) = (19/2)(52/5 + 18(7/5))
Sum(19) = (19/2)(52/5 + 126/5)
Sum(19) = (19/2)(178/5)
Sum(19) = (19/2)(178/5)
Sum(19) = (19 × 89) / 5
Sum(19) = 361
Therefore, the sum of the first 19 terms of the given AP is 361.
Let's first find the common difference (d) of the arithmetic progression (AP).
Given that the third term (a3) is 8, we can use the formula for the nth term of an AP:
an = a1 + (n - 1) * d
Substituting n = 3 and an = 8:
8 = a1 + (3 - 1) * d
8 = a1 + 2d
Now, let's use the information that the ninth term (a9) exceeds three times the third term by 2. We can set up the equation:
a9 = 3 * a3 + 2
Substituting a3 = 8:
a9 = 3 * 8 + 2
a9 = 24 + 2
a9 = 26
To find the sum of the first 19 terms of the AP, we can use the formula:
Sn = (n/2) * (a1 + an)
Substituting n = 19:
S19 = (19/2) * (a1 + a19)
To find a19, we can use the formula for the nth term of an AP:
a19 = a1 + (19 - 1) * d
a19 = a1 + 18d
Now we have:
S19 = (19/2) * (a1 + a1 + 18d)
S19 = (19/2) * (2a1 + 18d)
To find the sum, we need to find the common difference (d) and the first term (a1).
From the equations we derived:
8 = a1 + 2d
26 = a1 + 18d
We can solve these two simultaneous equations to find the values of a1 and d.
(26 - 8) = (a1 + 18d) - (a1 + 2d)
18 = 16d
d = 18/16
d = 9/8
Substituting d = 9/8 into the first equation:
8 = a1 + 2(9/8)
8 = a1 + 9/4
a1 = 8 - 9/4
a1 = (32 - 9)/4
a1 = 23/4
Now we can substitute a1 = 23/4 and d = 9/8 into the formula for S19:
S19 = (19/2) * (2(23/4) + 18(9/8))
S19 = (19/2) * (46/4 + 162/8)
S19 = (19/2) * (46/4 + 81/4)
S19 = (19/2) * (127/4)
S19 = (19 * 127) / (2 * 4)
S19 = 2413/8
Therefore, the sum of the first 19 terms of the AP is 2413/8.