Question

The 7th term of an AP is -39/12 and the 15th term is -103/12. What is the 27th term?
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please answer fast

Answer 1

Answer:

-199/12

Step-by-step explanation:

7th term = a + 6d

-39/12 = a + 6d ..... eq1

15th term = a + 14d

-103/12 = a + 14d ...... eq2

from eq1 & eq2

a = 3/4 & d = -2/3

so 27th term = a + 26d

27th term = -199/12

Answer 2

Answer:

$\bigstar{\bold{27th\:term=\dfrac{-199}{12} }}$

Step-by-step explanation:

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

• 7th term of an A.P is -39/12
• 15th term of an A.P is -103/12

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

• The 27th term

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

→ Here we have to find the first term and common difference of the A.P

→ The 7th term of an A.P is given by

   a₇ = a₁ + 6d

→ Substituting the datas we get

  -39/12 = a₁ + 6 d---(1)

→ The fifteenth term of an A.P is given by

   a₁₅ = a₁ + 14 d

→ Substituting the datas,

  -103/12 = a₁ + 14 d ----(2)

→ Solving equation 1 and 2 by elimination method

   a₁ + 14d = -103/12

   a₁ + 6d = -39/12

          8d = -64/12

            d = -8/12

            d = -2/3

→ Now substitute the value of d in equation 1

  a₁ + 6 × -2/3 = -39/12

  a₁ + -4 = -39/12

  a₁ = -39/12 + 4

  a₁ = 9/12

→ Hence the first term of the A.P is 9/12

→ Now 27th term of the A.P is given by

  a₂₇ = a₁ + 26 d

→ Substitute the values we get,

  a₂₇ = 9/12 + 26 × -2/3

  a₂₇ = 9/12 - 52/3

  a₂₇ = -199/12

→ Hence 27th term of the A.P is -199/12

$\boxed{\bold{27th\:term=\dfrac{-199}{12} }}$

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

→ nth term of an A.P is given by

  $\sf{a_=a_1+(n-1)\times d}$

→ The common difference of an A.P is given by

   $\sf{d=\dfrac{a_m-a_n}{m-n}$