Question

A man saves $Rs.200$ in each of the first three months of his service. In each of the subsequent months his savings increases by $Rs.40$ more than the savings of the previous month immediately. After how many months his total savings from the start of service will be $Rs.11040$.
(A) $18\;months$
(B) $21\;months$
(C) $28\;months$
(D) $10\;months$

Answer 1

Hint:We solve this questions by forming arithmetic progression. We consider terms as $200,200,200,240,280..........up\;to\;n\;terms$ according to information given in the question and then we making A.P we consider number of terms as of $(n - 2)$ terms as we will not consider the first two months saving for A.P. So, we apply the formula of the sum of n terms of an A.P. So we find the number of months we add the saving of the first two months in the sum of n terms of an A.P and find the value of n.

Complete step-by-step answer:
According to the question savings are $200,200,200,240,280..........up\;to\;n\;terms$.
But $200,240,280..........(n - 2)$ forms an A.P.
As we know that, sum of n terms of an A.P ${S_n} = 2n[2a + (n - 1)d]$,
Where n is the number of terms, a be the first term of the A.P and d is the common difference between the terms.
Now according to the question equation formed, we get,
$ \Rightarrow 400 + \dfrac{{n - 2}}{2}[2a + ((n - 2) - 1)d] = 11040$
So by putting the values in the equation we get,
$ \Rightarrow 400 + \dfrac{{n - 2}}{2}[2 \times 200 + (n - 2 - 1) \times 40] = 11040$
$ \Rightarrow 400 + \dfrac{{n - 2}}{2}[2 \times 200 + (n - 3) \times 40] = 11040$
By taking 2 common from the equation we get,
$ \Rightarrow 400 + 2(\dfrac{{n - 2}}{2})[200 + (n - 3) \times 20] = 11040$
Now by solving the equation we get,
$ \Rightarrow 400 + (n - 2)[200 + (n - 3) \times 20] = 11040$
$ \Rightarrow 400 + (n - 2)[200 + 20n - 60] = 11040$
We take $400$from L.H.s to the R.H.S, we get
$ \Rightarrow (n - 2)[200 + 20n - 60] = 11040 - 400$
$ \Rightarrow (n - 2)[200 + 20n - 60] = 10640$
After solving the L.H.S, we get
$ \Rightarrow (n - 2)[140 + 20n] = 10640$
Taking $20$ common from the equation, we get
$ \Rightarrow 20 \times (n - 2)[n + 7] = 10640$
We further solve the equation we get
$ \Rightarrow (n - 2)(n + 7) = \dfrac{{10640}}{{20}}$
$ \Rightarrow (n - 2)(n + 7) = 532$
We solve the equation multiply both the brackets and we perform addition of two negative terms, we get
$
   \Rightarrow {n^2} + 5n - 14 = 532 \\
   \Rightarrow {n^2} + 5n - 14 - 532 = 0 \\
   \Rightarrow {n^2} + 5n - 546 = 0 \\
 $
By applying splitting the middle term we get,
$ \Rightarrow {n^2} + (26 - 21)n + 26 \times ( - 21) = 0$
Now by solving the above equation we get
$ \Rightarrow {n^2} + 26n - 21n + 26 \times ( - 21) = 0$
By taking common values we get
$ \Rightarrow (n + 26)(n - 21) = 0$
$\because n = - 26$ is not possible because the value of the month is never negative
$\therefore n = 21$
So the total time will be $21\;months$

So, the correct answer is “Option B”.

Note:In these types of questions we should remember that the total number of terms must be considered as n and in A.P. we calculate it till $(n - 2)$ terms because he saves $Rs.200$ for three consecutive months. In other words we say that we do not consider the first two months saving in A.P during making the progression but remember that you should add the first two months saving while making the equation for finding the number of months.