**Q) ** In an AP, the sum of the first n terms is given by S_{n} = 6n – n^{2}. Find its 30^{th} term.

**Ans: **

We know that, the formula for the sum of the first n terms of an arithmetic progression (AP) is given by:

S_{n }_{ }= [2a + (n-1) d]

where:

- $Sn $ is the sum of the first n terms,
- $a$ is the first term,
- $d$ is the common difference,
- $n$ is the number of terms.

In this problem, we are given the sum formula as S_{n }$=6n−n2$. We can compare this with the formula for the sum of an AP:

∴ [2a + (2-1) d] = 6n – n^{2 }

∴ [2a + (n – 1) d] = n (6 – n)

∴ 2a + (n -1 ) d = 2 (6 – n) ………… (i)

Next, let’s find the value of first term a, Common difference d.

For ease of calculation, let’s take n = 2, equation (i) will result:

2a + (2 – 1) d = 2 (6 – 2)

2a + d = 8 …….. (ii)

Similarly, for n = 3, equation (i) will result:

2a + (3 – 1) d = 2 (6 – 3)

2a + 2d = 6 …….. (iii)

By solving equation (ii) and equation (iii), we get:

a = 5, d = – 2

Now, we have values of a and d, let’s calculate value of 30^{th} term:

We know that the n^{th} term of an AP is given by: T_{n } = a + (n – 1) d

Therefore, value of 30^{th} term, T_{30 }= (5) + (30 – 1) x (-2)

T_{30 }= 5 + 29 x – 2 = – 53

**Therefore, the value of 30 ^{th} term is – 53.**

**Note:** Students, There is another method which involves to first calculate S_{n-1 }and then calculate a_{n. } But in that method, the chances of calculation mistake are very high, that’s why, you should opt for calculating values by putting value of n. Again, smaller values of n should be taken to remove complexities of calculation. Here, your objective should be to score higher marks by doing precise solution, not to impress examiner.