How many terms of the AP: 24, 21, 18, ............. must be taken so…

The correct answer is Option (3) → Both 4 and 13

The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}\left[2a+(n-1)d\right]\), where \(a\) is the first term and \(d\) the common difference. Setting it equal to 78 gives a quadratic in \(n\), and both of its roots can be valid.

The first term and the common difference (each term is 3 less than the one before):

• \(a=24\)

• \(d=21-24=-3\)

The quadratic in \(n\) (putting \(a\) and \(d\) in the formula and setting the sum to 78):

• \(\displaystyle \frac{n}{2}\left[48+(n-1)(-3)\right]=78\)

• \(\displaystyle \frac{n}{2}(51-3n)=78\)

• So \(51n-3n^2=156\)

• \(3n^2-51n+156=0\), so \(n^2-17n+52=0\)

Its roots (4 and 13 multiply to 52 and add to 17):

• \((n-4)(n-13)=0\)

• \(n=4\) or \(n=13\)

Both are whole numbers, so both answers stand:

• 4 terms: \(24+21+18+15=78\)

• 13 terms: the 5th to 13th terms are 12, 9, 6, 3, 0, −3, −6, −9, −12, which add up to 0, so the sum is still 78