#### Question

Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

#### Solution 1

Given that,

*a*_{2} = 14

*a*_{3} = 18

*d* = *a*_{3} − *a*_{2} = 18 − 14 = 4

*a*_{2} = *a* + *d*

14 = *a* + 4

*a* = 10

`S_n = n/2[2a+(n-1)d]`

`S_51 = 51/2[2xx10+(51-1)4]`

`=51/2[20+(50)(4)]`

`=(51(220))/2=51(110)`

= 5610

#### Solution 2

In the given problem, let us take the first term as *a* and the common difference as d

Here, we are given that,

`a_2 = 14` .....(1)

`a_3 = 18` ....(2)

Also we know

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

For the 2^{nd} term (n = 2)

`a_2 = a + (2 - 1)d`

14 = a + d (Using 1)

a = 14 - d .....(3)

Similarly for the 3rd term (n = 3)

`a_3 = a + (3 - 1)d`

18 = a + 2d (Using 2)

a = 18 - 2d .......(4)

Subtracting (3) from (4), we get,

a - a = (18 - 2d) - (14 - d)

0 = 18 - 2d - 14 + d

0 = 4 - d

d = 4

Now, to find *a*, we substitute the value of *d* in (4),

a = 14 - 4

a = 10

So for the given A.P d = 4 and a = 10

So, to find the sum of first 51 terms of this A.P., we use the following formula for the sum of *n* terms of an A.P.,

`S_n = n/2 [2a + (n - 1)d]`

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

So, using the formula for *n* = 51, we get,

`S_51 = 51/2 [2(10) + (51 - 1)(4)]`

`= 51/2 [20 + (50)(4)]`

`= 51/2 [20 + 200]`

`= 51/2 [220]`

= 5610

Therefore the usm of first 51 terms for the given A.P is `S_51 = 5610`