Question

the third term of an A.P is 8 and the ninth term exceeds three times third by 2. find the sum of its first 19 terms

Answer 1

given,a+2d=8, a+8d=3(a+2d)+2
a+2d=8,. 2a+2d=2
2d=8-a,. 2a+8-a=2
a= -6
d=14
sum of 19 terms=19/2(2(-6)+18(14))
=19/2(240)=228

Answer 2

Answer:

551

Step-by-step explanation:

We are given $a_3= 8$   --1

Formula of nth term in A.P. = $a_n=a+(n-1)d$

Substitute n = 3

$a_3=a+(3-1)d$

$8=a+2d$  ---2  

Substitute n= 9

$a_9=a+(9-1)d$

$a_9=a+8d$  --3

Since we are given that the ninth term exceeds three times third by 2.

⇒$a_9 =3 a_3+2$

Using 1 and 3

⇒$a+8d =3 (8)+2$

⇒$a+8d =24+2$

⇒$a+8d =26$ --4

Solve 2 and 4 to find the value of a and d

Subtract 2 from 4

$a+8d-a-2d=26-8$

$8d-2d=18$

$6d=18$

$d=\frac{18}{6}$

$d=3$

Substitute the value of d in 2 to get value of a

$8=a+2(3)$

$8=a+6$

$2=a$

Thus a = first term = 2

d = common difference = 3

Formula of sum of first n terms :$S_n=\frac{n}{2}(2a+(n-1)d)$

Substitute n = 19

$S_{19}=\frac{19}{2}(2(2)+(19-1)(3))$

$S_{19}=\frac{19}{2}(4+54)$

$S_{19}=\frac{19}{2}(58)$

$S_{19}=551$

Hence the sum of first 19 terms is 551