Question

The third term of an ap is 8 and 9th term of an ap exceeds three times the third term by 2 . Find the sum of its first 19 terms

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Assumption

First term be a

Common difference be d

${\boxed{\sf\:{Hence}}}$

nth term :-

an = a + (n - 1)d

Given,

${\boxed{\sf\:{a_{3}=a+(n-1)d}}}$

a + (3 - 1)d

a + 2d = 8 .....(1)

We have

${\boxed{\sf\:{9th\;term\;exceeds\;3\;times\;the\;third\; term\;by\;2}}}$

Hence,

${\boxed{\sf\:{a_{9}=3a_{3}+2}}}$

a + (9 - 1)d = 3(8) + 2 [a3 = 8]

a + 8d = 26 .......(2)

${\boxed{\sf\:{Subtract\;(1)\;from\;(2)}}}$

6d = 18

${\boxed{\sf\:{d=\dfrac{18}{6}}}}$

d = 3

${\boxed{\sf\:{Substitute\;value\;of\;d\;in\;(1)}}}$

a + 2d = 8

a + 2 × 3 = 8

a + 6 = 8

a = 8 - 6

a = 2

${\boxed{\sf\:{Sum\;of\;19th\;term}}}$

${\boxed{\sf\:{S_{n}=\dfrac{n}{2}[2a+(n-1)d]}}}$

$\tt{\rightarrow S_{19}=\dfrac{19}{2}[2\times 2+(19-1)3]}$

$\tt{\rightarrow S_{19}=\dfrac{19}{2}[4+18\times 3]}$

$\tt{\rightarrow S_{19}=\dfrac{19}{2}[4+54]}$

$\tt{\rightarrow S_{19}=\dfrac{19}{2}[58]}$

$\tt{\rightarrow S_{19}=\dfrac{1102}{2}}$

= 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.