Question & Answer
QUESTION

Let ${a_1},{a_2},{a_3},....,{a_{100}}$ be arithmetic progression with ${a_1} = 3$ and ${S_p}$ is sum of 100 terms. For any integer n with $1 \leqslant n \leqslant 20$ , let $m = 5n$ . If $\dfrac{{{S_m}}}{{{S_n}}}$ does not depend on n, then ${a_2}$ is
A. 6
B. 7
C. 8
D. 9

ANSWER Verified Verified
Hint: Here we will use the sum of first p terms formula of an Arithmetic Progression i.e. $[{S_p} = \dfrac{p}{2}[2a + (p - 1)d]$ to calculate the common difference(d).

Complete step-by-step answer:
 We know that, if the first term of the AP is a and common difference is d then the sum of first x terms can be found by the formula ${S_p} = \dfrac{p}{2}[2a + (p - 1)d]$ . Using this formula,
\[
  \dfrac{{{S_m}}}{{{S_n}}} = \dfrac{{{S_{5n}}}}{{{S_n}}} = \dfrac{{\dfrac{{5n}}{2}[2 \times 3 + (5n - 1)d]}}{{\dfrac{n}{2}[2 \times 3 + (n - 1)d]}}{\text{ }}[Given,m = 5n] \\
   \Rightarrow \dfrac{{{S_{5n}}}}{{{S_n}}} = \dfrac{{5n[6 + (5n - 1)d]}}{2} \times \dfrac{2}{{n[6 + (n - 1)d]}} \\
   \Rightarrow \dfrac{{{S_{5n}}}}{{{S_n}}} = \dfrac{{30 + 5d(5n - 1)}}{{6 + (n - 1)d}} \\
   \Rightarrow \dfrac{{{S_{5n}}}}{{{S_n}}} = \dfrac{{30 + 25dn - 5d}}{{6 + dn - d}} \\
   \Rightarrow \dfrac{{{S_{5n}}}}{{{S_n}}} = \dfrac{{5(6 - d) + 25dn}}{{(6 - d) + dn}} \\
 \]
Hence, we can conclude that for $\dfrac{{{S_m}}}{{{S_n}}}$ to be independent from n, d has to be 6.
But we need to find ${a_2}$ with this condition
.
 So, ${a_2} = {a_1} + d = 3 + 6 = 9$ . Hence, the value of ${a_2}$ with a given condition is 9.
Answer is option D.

Note: In any question of AP, if we can get its first term and common difference then we can get anything whatsoever the question is asking. That’s what we did in this problem. First, we found the value of d. Then we calculated ${a_2}$. Student should remember the formula of sum of n terms of A.P and general form of A.P.