Question

The sum of 4th and 8th terms of an A.P. is 24 and the sum of 6th and 10th terms is 44. Find the A.P.

Answer 1

In the given problem, the sum of 4^th and 8^th term is 24 and the sum of 6^th and 10^thterm is 44. We have to find the A.P

We can write this as

`a_4 + a_8 = 24` ......(1)

`a_6 + a_10 = 44` .....(2)

We need to find the A.P

For the given A.P., let us take the first term as a and the common difference as d

As we know,

`a_n = a + (n - 1)d`

For 4^th term (n = 4)

`a_4 = a + (4 - 1)d`

= a + 7d

So, on substituting the above values in (1), we get,

(a + 3d) + (a + 7d) = 24

2a + 10d = 24  ....(3)

Also for 6^th term (n = 6)

`a_6 = a + (6 - 1)d`

= a + 5d

For 10^th term (n = 10)

`a_10 = a + (10 - 1)d`

= a + 5d

For 10 ^th term (n = 10)

`a_10 = a + (10 -1)d`

= a + 9d

So on substituting the above values in 2 we get

`(a + 5d) + (a + 9d) = 44`

2a + 14d = 44 .......(4)

Next we simplify (3) and (4). On subtracting (3) from (4), we get,

`(2a + 14d)- (2a + 10d) = 44 - 24`

2a + 14d - 2a - 10d= 20

4d = 20

`d = 20/4`

d = 5

Further, using the value of d in equation (3), we get,

a + 10(5) = 24

2a + 50 = 24

2a = 24 - 50

2a = -26

a = -13

So here a= -13 and d = 5

Therefore the A.P is -13, -8, -3, 2,.....