Question

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

Answer 1

Answer:

The sum of first 19 terms is 551.

Step-by-step explanation:

a + 2d = 8.____(1)

a + 8d - 3 x 8 = 2

a + 8d = 26____(2)

Subtract 1 from 2 :

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

6d = 18

d = 3.

Now,

a + 2d = 8.

a + 2 x 3 = 8

a + 6 = 8

a = 2.

Now,

Sum of First 19 terms = 19/2(2 x 2 + 18 x 3 )

Sum of first 19 terms = 19/2( 4 + 54)

Sum of first 19 terms = 19/2 x 58

Sum of first 19 terms = 551

Answer 2

The nth term of an A.P with first term a and common difference d is

T n ​ =a+(n−1)d.

Here, it is given that the third term of an A.P is 8, therefore,

⇒T 3 ​ =a+(3−1)d

⇒8=a+2d

⇒a+2d=8...…(1)

It is also given that the ninth term of an A.P exceeds three times the third term by 2, therefore,

⇒T 9 ​ =3T 3 ​ +2=(3×8)+2=24+2=26

But ⇒T 9 ​ =a+(9−1)d=a+8d, thus,

⇒a+8d=26...…(2)

Now, subtract equation 1 from equation 2 as follows:

⇒(a−a)+(8d−2d)=26−8

⇒6d=18

⇒d= 6 18 ​ =3

Substitute d=3 in equation 1:

a+(2×3)=8

⇒a+6=8

⇒a=8−6=2

We also know that the sum of n terms of an A.P with first term a and common difference d is:

⇒S n ​ = 2 n ​ [2a+(n−1)d]

⇒Substitute n=19, a=2 and d=3 in S n ​ = 2 n ​ [2a+(n−1)d] as follows:

⇒S 19 ​ = 2 19 ​ [(2×2)+(19−1)3]

= 2 19 ​ [4+(18×3)]

= 2 19 ​ (4+54)

= 2 19 ​ ×58

=19×29

=551

Hence, the sum of the first 19 terms of an A.P is S 19 ​ =551.