There are (n - 1) sets prior to set n and those sets collectively contain Σr elements where r = 1 to n - 1 i.e. n(n - 1)/2 elements.

Hence, the first term in set n is the n(n - 1)/2 + 1 term of the AP of first term 2 and common difference 4. This is equivalent to 2 + [n(n - 1)/2 + 1 - 1] × 4 = 2n² - 2n + 2.

There are n terms in the n set. Hence, the last term in the n set is given by 2n² - 2n + 2 + (n - 1) × 4 = 2n² + 2n - 2.

Additionally, the sum of set n is given by n/2 (2n² - 2n + 2 + 2n² + 2n - 2) = 2n³.

There are n terms in the nth set.

Number of terms in first n sets = 1 + 2 + 3 + ... + n = n(n+1)/2
(Sum of AP)

The last term has the same numbering as the number of terms.
It is the n(n+1)/2 th term.

The progression has first term 2 and common difference 4.

Last term = 2 + 4 × (n(n+1)/2 - 1)

= 2 + 2n² + 2n - 4
= 2n² + 2n - 2

This answers ii)

For the first term in the nth set, just subtract (n - 1) times of 4.

2n² + 2n - 2 - 4(n - 1)

= 2n² + 2n - 2 - 4n + 4
= 2n² - 2n + 2

This answers i)


To answer iii), use the sum of AP on the last set.

(First term in the nth set + last term in the nth set) × (number of terms in the nth set) ÷ 2


(2n² - 2n + 2 + (2n² + 2n - 2)) × n ÷ 2

= 4n² × n ÷ 2
= 2n³

Alternative for ii) :

You can also find the '0th' term first. Then add n(n+1)/2 times of 4 to it.

'0th' term = 2 - 4 = -2

-2 + 4 × n(n+1)/2

= -2 + 2n² + 2n

(Rewrite as 2n² + 2n - 2)