Q)Â Â In an AP, the sum of the first n terms is given by Sn = 6n – n2. Find its 30th term.
Ans:
VIDEO SOLUTION
STEP BY STEP SOLUTION
Step 1: We know that, the formula for the sum of the first n terms of an arithmetic progression (AP) is given by:
Sn  =
[2a + (n-1) d]
where:
- is the sum of the first n terms,
- is the first term,
- is the common difference,
- is the number of terms.
In this problem, we are given the sum formula as Sn .
Step 2: Let’s compare this with the formula for the sum of an AP:
∴
[2a + (2-1) d] = 6n – n2Â
∴
[2a + (n – 1) d] = n (6 – n)
∴ 2a + (n -1 ) d = 2 (6 – n) ………… (i)
Step 3: Next, let’s find the value of first term a, Common difference d.
For ease of calculation, let’s take n = 1, equation (i) will result:
2a + (1 – 1) d = 2 (6 – 1)
2a + 0 = 10
a = 5 …….. (ii)
Step 4: Next for n = 2, equation (i) will result:
2a + (2 – 1) d = 2 (6 – 2)
2a + d = 8
Now let’s substitute a = 5 from equation (ii):
2 (5) + d = 8
10 + d = 8
d = – 2 ….. (iii)
So now from equations (i) & (ii), we have: a = 5, d = – 2
Step 5: Now, let’s calculate value of 30th term:
We know that the nth term of an AP is given by: Tn  = a + (n – 1) d
Therefore, value of 30th term, T30 = (5) + (30 – 1) x (-2)
T30 = 5 + 29 x – 2 = – 53
Therefore, the value of 30th term is – 53.
Note: Students, There is another method which involves to first calculate Sn-1 and then calculate an. Â 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.
