Question

In an ap whose first terms, third term, and the fifth terms is 39 and the sum of second term, fourth and the sixth term is 51, find the tenth term of the sequence

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

Given; The sum of first term, third term and fifth term is 39 & the sum of second, fourth and sixth terms is 51

To find: The 10th term of the A.P

Solution:

The sum of first term, third term and fifth term is 39

i.e. Let the first term be ' a '

Third term be ' a + 2d '

Fifth term be ' a + 4d '

Sum = 39 ( given )

→ a + a + 2d + a + 4d = 39

→ 3a + 6d = 39

→ 3 ( a + 2d ) = 39

→ a + 2d = 13 -----[1]

The sum of second, fourth and sixth terms is 51

Let the second term be ' a + d '

Fourth term be ' a + 3d '

Sixth term be ' a + 5d '

Sum = 51 ( given )

→ a + d + a + 3d + a + 5d = 51

→ 3a + 9d = 51

→ 3 ( a + 3d ) = 51

→ a + 3d = 17 -----[2]

Solve equation [1] & [2]

We get,

$a + 2d = 13 \: -----[1]\\\underline{a + 3d = 17} \: -----[2]\\-d = -4$

$\underline{\boxed{\bold{D = 4}}}$

Substitute 'd' in [1] or [2]

→ a + 3d = 17

→ a + 12 = 17

→ a = 17 - 12 = 5

10'th term = a + 9d

→ a + 9d

→ 5 + 36 = 41

⇒ a₁₀ 41

$\underline{\boxed{\bold{a_{10} = 41}}}$