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Question: In another case , p and 2p are the first and second terms of an arithmetic progression. The nth term...

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

Explanation

Solution

With the given terms we can find the common difference and using the nth term formula an=a+(n1)d{a_n} = a + (n - 1)d we can find a equation substituting in the formula of Sn=n2[2a+(n1)d]{S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]. we can find the value of n and p.

Complete step by step solution:
We are given that the first two terms of an AP are p and 2p
a=p,a2=2p\Rightarrow a = p,{a_2} = 2p
The common difference is given by subtracting the first term from the second term
\Rightarrow 2p – p = p
The nth of the sequence is given by
an=a+(n1)d\Rightarrow {a_n} = a + (n - 1)d
Here we have the nth term to be 336
336=p+(n1)p 336=p+pnp=pn np=336  \Rightarrow 336 = p + (n - 1)p \\\ \Rightarrow 336 = p + pn - p = pn \\\ \Rightarrow np = 336 \\\
We know that the sum of the first n terms is given by
Sn=n2[2a+(n1)d]\Rightarrow {S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]
7224=n2[2p+(n1)p] 7224=n2[2p+npp] 7224=n2[p+np]   \Rightarrow 7224 = \dfrac{n}{2}\left[ {2p + (n - 1)p} \right] \\\ \Rightarrow 7224 = \dfrac{n}{2}\left[ {2p + np - p} \right] \\\ \Rightarrow 7224 = \dfrac{n}{2}\left[ {p + np} \right] \\\ \\\
Substituting the value of np we get
7224=n2[p+336] 7224=np2+336n2 7224=3362+168n 7224=168+168n 7224=168(1+n) 7224168=n+1 43=n+1 n=42  \Rightarrow 7224 = \dfrac{n}{2}\left[ {p + 336} \right] \\\ \Rightarrow 7224 = \dfrac{{np}}{2} + \dfrac{{336n}}{2} \\\ \Rightarrow 7224 = \dfrac{{336}}{2} + 168n \\\ \Rightarrow 7224 = 168 + 168n \\\ \Rightarrow 7224 = 168(1 + n) \\\ \Rightarrow \dfrac{{7224}}{{168}} = n + 1 \\\ \Rightarrow 43 = n + 1 \\\ \Rightarrow n = 42 \\\
Substituting this in np = 336
42p=336 p=33642=8  \Rightarrow 42p = 336 \\\ \Rightarrow p = \dfrac{{336}}{{42}} = 8 \\\

Therefore the value of p = 8 and n = 42

Note:

  1. In an Arithmetic Sequence the difference between one term and the next is a constant.
  2. We can find the common difference of an AP by finding the difference between any two adjacent terms.
  3. If we know the initial term, the following terms are related to it by repeated addition of the common difference.