Question

in an arithmetic progression the sum of the first term the third term and the fifth term is 39 find the sum of second term fourth term and sixth term is 51 find the 10th term of the sequence​

Answer 1

Answer:

A penalty of an extra that is conceded by the bowler if he breaks the rules of bowling either by using an inappropriate arm action. Overstepping popping crease or Having a foot outside return crease.

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

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