Question

Let \[{{a}_{1}},{{a}_{2}},{{a}_{3}},.....{{a}_{100}}\] be an arithmetic progression with \[{{a}_{1}}=3\] and \[{{S}_{p}}\] is sum of 300. For any integer n with \[1\le n\le 20\], let \[m=5n\]. If \[\dfrac{{{S}_{m}}}{{{S}_{n}}}\] does not depend on n, then \[{{a}_{2}}\].
1. 9
2. 2 or 4
3. 4 or 16
4. None of these

Answer 1

Hint: In this particular problem first we need to recall the formula for sum of series which is in AP
Because the series which is given is in arithmetic progression. So, the formula for arithmetic progression is given by \[{{S}_{k}}=\dfrac{k}{2}(2a+(k-1)d)\]. After taking the ratio of \[\dfrac{{{S}_{m}}}{{{S}_{n}}}\] we get the value of d as it is independent of n. so, in this way we have to solve further and get the answer.

Complete step-by-step solution:
In this particular problem, it has mentioned in the series given that is \[{{a}_{1}},{{a}_{2}},{{a}_{3}},.....{{a}_{100}}\]
Above series is given in the form of A.P. so we have to apply the formula for the sum of series in A.P.
\[{{S}_{k}}=\dfrac{k}{2}(2a+(k-1)d)\]
Here, in this question it has also given the relation between m and n.
That is \[m=5n\] where, ratio \[\dfrac{{{S}_{m}}}{{{S}_{n}}}\] is not dependent on n.
First of all we need to find the value of \[{{S}_{m}}\]
\[{{S}_{m}}=\dfrac{m}{2}(2a+(m-1)d)\]
But we know that \[m=5n\] substitute in above equation that is
\[{{S}_{m}}=\dfrac{5n}{2}(2a+(5n-1)d)---(1)\]
Now, we have to find the value of \[{{S}_{n}}\]
\[{{S}_{n}}=\dfrac{n}{2}(2a+(n-1)d)---(2)\]
We divide the equation (1) and equation (2)
\[\dfrac{{{S}_{m}}}{{{S}_{n}}}=\dfrac{\dfrac{5n}{2}(2a+(5n-1)d)}{\dfrac{n}{2}(2a+(n-1)d)}\]
Here in this equation you can substitute the value of \[a=3\] we get:
\[\dfrac{{{S}_{m}}}{{{S}_{n}}}=\dfrac{\dfrac{5n}{2}(2(3)+(5n-1)d)}{\dfrac{n}{2}(2(3)+(n-1)d)}\]
After simplifying this above equation we get:
\[\dfrac{{{S}_{m}}}{{{S}_{n}}}=\dfrac{5n(6+(5n-1)d)}{n(6+(n-1)d)}\]
Here, n gets cancelled we get:
\[\dfrac{{{S}_{m}}}{{{S}_{n}}}=\dfrac{5(6+(5n-1)d)}{(6+(n-1)d)}\]
After simplifying further we get:
\[\dfrac{{{S}_{m}}}{{{S}_{n}}}=\dfrac{5((6-d)+5nd)}{((6-d)+nd)}---(3)\]
Condition which is given that
\[\dfrac{{{S}_{m}}}{{{S}_{n}}}\] Does not depend on n.
Hence, \[d=6\]
Because if we substitute \[d=6\] equation (3) becomes independent on n that means the expression is not in the term of n.
As we know that common difference should be the same for A.P.
\[{{a}_{2}}-{{a}_{1}}=d\]
Here, \[d=6\]and \[{{a}_{1}}=3\]
\[\therefore {{a}_{2}}-3=6\]
\[\therefore {{a}_{2}}=6+3\]
\[\therefore {{a}_{2}}=9\]
So, the correct option is “option 1”.

Note: In this particular problem, we have to keep in mind that the formula which we used is important. And don’t make silly mistakes while substituting the values. Another important thing is that when we are calculating the value of d. Always focus the statement which is given in question that is \[\dfrac{{{S}_{m}}}{{{S}_{n}}}\] does not depend on n that means they have indirectly written here that we have to select the value of d in such a way the expression should be free of n. If we substitute the value of \[d=6\] then in the expression n is cancelled and the ratio is 25 that means the ratio is not in the term of n. That we have to remember always, and don’t make silly mistakes here while selecting the values of d. So, the above solution is preferred for such types of problems.