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

Here is your answer.....

Answer 2

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

an = a + (n - 1)d

Third term = 8.

Hence

⇒ a + 2d = 8 ......(Eqn 1)

9th term exceeds 3 times the third term by 2.

Therefore, 

a9 = 3a3 + 2

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

⇒ a + 8d = 26 .........(Eqn 2)

Subtracting (1) from (2) we get

⇒ 6d = 18

$\bf\huge{\implies d = \dfrac{18}{6}}$

⇒ d = 3

Putting the value of d in Equation (1),

⇒ a + 2d = 8

⇒ a + 2 × 3 = 8

⇒ a + 6 = 8

⇒ a = 8 - 6

⇒ a = 2

Hence

$\bf\huge{\implies S_{n}=\dfrac{n}{2}[2a + (n - 1)d]}$        

Sum of first 19 terms

$\bf\huge{\implies S_{19}=\dfrac{19}{2}[2\times 2 + (19 - 1)3]}$  

$\bf\huge{\implies S_{19} = \dfrac{19}{2}[4 + 18\times 3]}$  

$\bf\huge{\implies S_{19} = \dfrac{19}{2}[4 + 54]}$

$\bf\huge{\implies S_{19} = \dfrac{19}{2}[58]}$

$\bf\huge{\implies\dfrac{1102}{2}}$

= 551

$\bf\huge\bf\huge{\boxed{\bigstar{{Sum\: of\: the\: first\: 19\: term = 551}}}}$