Question

find the m th term of an arithmatic progression whose 12 th term exceeds the 5 th term by 14 and the sum of both term is 36.​

Answer 1

Answer:

$\bigstar{\bold{Mth\:term=2m+1}}$

Step-by-step explanation:

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

• The 12th term of the A.P exceeds the 5th term by 14
• Sum of both the term is 36

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

• The mth term

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

➢ First we have to find the common difference and first term of the A.P

➢ The 12 th term of an A.P is given by,

     a₁₂ = a₁ + 11d

➢ The 5th term of an A.P is given by,

     a₅ = a₁ + 4d---------(1)

➢ By given,

    a₁₂ = a₅ + 14

➢ Substitute the value of a₅ from equation 1

➢ Hence,

    a₁ + 11d = a₁ + 4d + 14

➢ Cancelling a₁ on both sides,

    11d = 4d + 14

    11d - 4d = 14

     7d = 14

       d = 14/7

       d = 2

➢ Hence common difference of the A.P is 2.

➢ Now also by given,

    a₅ + a₁₂ = 36

    a₁ + 4d + a₁ + 11d = 36

    2a₁ + 15d = 36

➢ Substitute the value of d,

    2a₁ + 15 × 2 = 36

     2a₁ = 6

       a₁ = 3

➢ Hence the first term of the A.P is 3.

➢ Now the mth term of the A.P is given by,

    $\sf{a_m=a_1+(m-1)\times d}$

➢ Substitute the values,

     $\sf{a_m=3+(m-1)2}\\a_m=3+2m-2\\a_m=2m+1$

➢ Hence the mth term of the A.P is 2m + 1