The third term of an A.P is 8 and the ninth term of an A.P exceeds three times the term by 2 find the sum of its first 19 terms
santripti1234567890:
exceeds 3 times the which term?
May be 3rd term
Because they mentioned before about the third term
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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...........
Hope it helps.....
Mark as brainiest...........
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...........
mark as brainliest
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