#### Question

The sum of 4^{th}^{ }and 8^{th} terms of an A.P. is 24 and the sum of the 6^{th} and 10^{th} terms is 34. Find the first term and the common difference of the A.P.

#### Solution

In the given problem, the sum of 4^{th} and 8^{th} term is 24 and the sum of 6^{th} and 10^{th}term is 34.

We can write this as,

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

`a_6 + a_10 = 34` ......(2)

We need to find *a* and *d*

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 + 3d

For 8th term (n = 8)

`a_8 = a + (8 - 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 6th term (n = 6)

`a_6 = a + (6 - 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)= 34

2a + 14d = 34 ......(4)

Next we simplify 3 and 4. On substracting 3 from 4 we get

(2a + 14d) - (2a + 10d) = 34 - 24

2a + 14d - 3a - 10d = 10

4d =10

`d = 10/4`

d = 5/2

Further using the value of d in equation 3 we get

`a + 10(5/2) = 24`

2a + 5(5) = 24

2a = 24 - 25

On furthur simplifying we get

2a = -1

`a = (-1)/2`

Therefore for the given A.P `a = (-1)/2` and `d = 5/2`