Question

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

Answer 1

Answer:

$\bigstar{\bold{Sum\:of\:19\:terms=551}}$

Step-by-step explanation:

$\Large{\underline{\rm{Given:}}}$

• The third term of an A.P = 8
• Ninth term of the A.P = 3 times the third term by 2

$\Large{\underline{\rm{To\:Find:}}}$

• Sum of first 19 terms

$\Large{\underline{\rm{Solution:}}}$

➟ We know that the third term of the A.P = 8

    a₁ + 2d = 8 -----(1)

➟ Also by given we know that,

    Ninth term of the A.P (a₉) = 3 × (Third term) + 2

     a₉ = 3 × 8 + 2

     a₉ = 24 + 2

     a₉ = 26

➟ Also,

   a₁ + 8d = 26 ------(2)

➟ Solving equation 1 and equation 2 by elimination method,

   a₁ + 8d = 26

   a₁ + 2d = 8

          6d = 18

            d = 18/6

            d = 3

➟ Hence common difference of the A.P is 3.

➟ Now substitute the value of d in equation 1,

    a₁ + 2 × 3 = 8

    a₁ + 6 = 8

    a₁ = 8 - 6 =2

➟ Hence first term of the A.P is 2 .

➟ Now sum of terms of an A.P is given by,

    $\tt{S_n=\dfrac{n}{2} (2a_1+(n-1)\times d)}$

   where n = number of terms

               a₁ = first term

              d = common difference

➟ Substitute the given data,

    S₁₉ = 19/2 ( 2 × 2 + (19 - 1) × 3)

    S₁₉ = 9.5 × (4 + 54)

    S₁₉ = 551

➟ Hence the sum of 19 terms of the A.P is 551.