Question

In another case, p and 2p are the first and second term respectively of an arithmetic progression. The nth term is 336 and the of the first n terms is 7224. Write down two equations in n and p and hence find the values of n and p​

Answer 1

Consecutive terms in this sequence differ by p.

First term: p

Second term: p + p = 2p

Third term: 2p + p = 3p

and so on. It follows that the n-th term satisfies

np = 336

Presumably you meant to say the "sum of the first n terms" is 7224, which is to say

p + 2p + 3p + … + np = 7224

which can be rewritten as

p (1 + 2 + 3 + … + n) = 7224

p (n (n + 1)/2) = 7224

n (n + 1) p = 14,448

Substitute the first equation in the second one and solve for n :

336 (n + 1) = 14,448

n + 1 = 43

n = 42

Now solve for p :

42p = 336

p = 8