Q)   In an AP, the sum of the first n terms is given by Sn = 6n – n2. Find its 30th term.

Ans:

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 . We can compare this with the formula for the sum of an AP:

[2a + (2-1) d] = 6n – n

[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 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.  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.

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