Q)   In an AP, the sum of the first n terms is given by S_n = 6n – n². 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:

• S_n ​ 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 – n² 

∴ [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, T₃₀ = (5) + (30 – 1) x (-2)

T₃₀ = 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.