Question

The 3 rd of the AP is 8 and the ninth term of the AP exceeds three times the 3 rd term by 2.Find the sum of its first 19 th term

Answer 1

Answer:

Sum = 551

Step-by-step explanation:

t3 = a+2d = 8 ------(1)

t9 = a+8d

t3 = a+2d

a+8d = 3(a+2d) + 2

a+8d = 3a+6d + 2

3a + 6d - a - 8d = -2

2a - 2d = -2 -----(2)

a -d = -1 ------ (2)

By simultaneous equations

a = 2

d = 3

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

S19 = 19/2 * (2*2+(18)3) = 9.5 * (4+54) = 9.5 * 58 = 551

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Answer 2

an=a+(n-1)d
a3=a+(3-1)d
a3=a+2d
a3=8
a+2d=8...........1
an=a+(n-1)d
a9=a+(9-1)d
a9=a+8d
a9=8+8×3=32
a+8d=32..........2
2and1 we get
a+2d-(a+8d)=8-32
a+2d-a-8d=-24
-6d=-24
d=-24/-6

d=4
put ..1 to d
a+2d=8
a+2(4)=8
a=8-8
a=0
sn=n/2[2a+(n-1)d
s19=19/2[2×0+(19-1)4]
=19/2(0+18×4)
=(19/2)×72
=(19×36)
=684
than the sum of its first 19th term is 684