Question

in an AP the sum of first term ,third term and fifth term is 39 and sum of second term, 4th term and 6th term is 51. find the 10th term of the sequence

Answer 1

This may surely help you

Answer 2

Let the first term be a and the common difference be d

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

⇒ a + ( a + 2d) + (a + 4d) = 39

⇒ a + a + 2d + a + 4d = 39

⇒ 3a + 6d = 39

⇒ a + 2d = 13 --------------------[ 1 ]

The sum of second term, 4th term and 6th term is 51

⇒ ( a + d) + (a + 3d) + (a + 5d) = 51

⇒  a + d + a + 3d + a + 5d = 51

⇒ 3a + 9d = 51

⇒ a + 3d = 17 --------------------[ 2 ]

Put the two equations together:

a + 2d = 13 --------------------[ 1 ]

a + 3d = 17 --------------------[ 2 ]

Equation [ 2 ] - [ 1 ] :

d = 4

Find a:

Sub d = 4  into equation 1

a + 2(4) = 13

a + 8 = 13

a = 5

Form the nth term:

an = a1 + (n - 1)d

an = 5 + (n - 1)4

an = 5 + 4n - 4

an = 4n + 1

Find the 10th term:

an = 4n + 1

a10 = 4(10) + 1

a10 = 41

Answer: The 10th term is 41