Question

If \[{S_n}\] denotes the sum of first n terms of an A.P. whose first term is a and \[\dfrac{{{S_{nx}}}}{{{S_x}}}\] is independent of x, then \[{S_p} = \]
A) \[{P^3}\]
B) \[{P^2}a\]
C) \[P{a^2}\]
D) \[{a^3}\]

Answer 1

Hint: In this question we have to find the sum of $p$ term. Apply formula of sum of $n$ terms of AP to find ration of sum of $nx$ term to the sum of $x$ term and use given property to find relation between first term and common difference.

Formula used: \[{S_n} = \dfrac{n}{2}(2a + (n - 1)d)\]
Where
\[{S_n}\]Sum of n terms of AP
n is number of terms
a is first term
d is common difference

Complete step by step solution: Given: ratio of sum of $nx$ terms to the sum of $x$ terms are independent of $x$
\[\dfrac{{{S_{nx}}}}{{{S_n}}} = \dfrac{{\dfrac{{nx}}{2}[2a + (nx - 1)d]}}{{\dfrac{x}{2}[2a + (x - 1)d]}}\]
\[\dfrac{{{S_{nx}}}}{{{S_n}}} = \dfrac{{n[(2a - d) + nxd]}}{{(2a - d) + xd}}\]
It is given in the question that \[\dfrac{{{S_{nx}}}}{{{S_x}}}\]ratio is independent of x therefore
\[2a - d = 0\]
\[2a = d\]
Now sum of p terms is given as
\[{S_n} = \dfrac{n}{2}(2a + (n - 1)d)\]
\[{S_p} = \dfrac{p}{2}(2a + (p - 1)d)\]
We know that \[2a = d\]
\[{S_p} = \dfrac{p}{2}(2a + (p - 1)2a)\]
\[{S_p} = {p^2}a\]

Thus, Option (B) is correct.

Note: Here in this question only first term is given and common difference is unknown. So in order to find sum of AP either common difference, first term is known or relation between common difference and first term is known.
In order to find relation between common difference and first term always follow the condition which is given in question. Don’t try any other concept to find it otherwise it will become very complicated.
Sometime students get confused in between AP and GP the only difference in between them is in AP we talk about common difference whereas in GP we talk about common ratio.