Question

In an AP, the first term is 2, the last term is 29 and the sum of all the terms is 155. Find the common difference.

Answer 1

Here , a = 2 , l= 29   and s_n = 155 

Let d be the common difference of the given AP and n be the total number of terms.
Then, `T_n = 29`

⇒ a + (n-1) d = 29 

⇒ 2 +(n-1) d= 29          ..............(1)

The sum of n terms of an AP is given by

`s_n = n/2 [ a+l] = 155`

`⇒  n/2 [ 2 +29 ] = (n/2) xx 31 = 155`

 ⇒  n = 10

Putting the value of n in (i), we get:

⇒ 2 +9d = 29 

⇒ 9d = 27

⇒ d= 3 

Thus, the common difference of the given AP is 3.

 

Answer 2

In the given problem, we have the first and the last term of an A.P. along with the sum of all the terms of A.P. Here, we need to find the common difference of the A.P.

Here,

The first term of the A.P (a) = 2

The last term of the A.P (l) = 29

Sum of all the terms (S_n) = 155

Let the common difference of the A.P. be d.

So, let us first find the number of the terms (n) using the formula,

    155 = `(n/2) (2+29)`

    155 = `(n/2) (31)`

155(2) = (n)(31)

      `n = 310/31`

       n = 10

Now, to find the common difference of the A.P. we use the following formula,

l = a +(n-1) d

We get,

     29 = 2 + (10 - 1 ) d

     29 = 2 + (9)d

29 - 2 = 9d

      `d = 27/9`

       d = 3

Therefore, the common difference of the A.P. is d = 3 .