Question

The 7th term of an AP is 39/12 and the 15th term is -103/12what is the 27th term

Answer 1

Answer:-

Given:-

7th term of an AP (a₇) = - 39/12

15th term (a₁₅) = - 103/12

We know that,

nth term of an AP (aₙ) = a + (n - 1)d

Hence,

★ a₇ = a + (7 - 1)d

⟹ - 39/12 = a + 6d -- equation (1)

Similarly;

★ - 103/12 = a + 14d -- equation (2)

subtract equation (1) from (2).

⟹ a + 14d - (a + 6d) = - 103/12 - ( - 39/12)

⟹ a + 14d - a - 6d = ( - 103 + 39)/12

⟹ 8d = - 64/12

⟹ d = (- 64/12) * 1/8

⟹ d = - 64/96

⟹ d = - 2/3

Substitute d = - 2/3 in equation (1).

⟹ - 39/12 = a + 6( - 2/3)

⟹ - 13/4 = a - 4

⟹ - 13/4 + 4 = a

⟹ ( - 13 + 16)/4 = a

⟹ 3/4 = a

Now,

⟹ a₂₇ = a + 26d

Putting the respective values we get,

⟹ a₂₇ = 3/4 + 26(- 2/3)

⟹ a₂₇ = (9 - 208)/12

⟹ a₂₇ = - 199/12

∴ The 27th term of the given AP is - 199/12.

Answer 2

Answer:

Correct Question :-

• The 7th term of an AP is 39/12 and the 15th term is -103/12what is the 27th term.

Given :-

• The 7th term of an AP is 39/12 and the 15th term is -103/12.

Find Out :-

• What is the 27th term.

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

Henceforth, 27th term of the A.P is -199/12.

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