Question

Find the middle term of the sequence formed by all three digit numbers which leave a remainder 3 when divided by 4, also find the sum of all numbers on both sides of the middle term separately.

Answer 1

Hint – All the 3 digits numbers start from 100 and 100 when divided by 4 leaves the remainder 0, but the remainder should be 3 so add 3 to 100 hence the starting of the series will be from 103, now keep on adding 4 to the previous number. We can see that the series formed is in arithmetic progression so use the respective formula of the series to get the answer.

Complete step-by-step answer:

As we know, a three digit number is starting from 100.

And we all know 100 is divided by 4.

So the first three digit number when divided by 4 leaves the remainder 3 should be (100 + 3) = 103.

So the next three digit number when divided by 4 leaves the remainder 3 is (103 + 4) = 107.

So the series of 3 digit numbers when divided by 4 leaves the remainder 3 are
103, 107, 111, ………...., 999

So as we see that the above series forms an A.P with first term (a1) = 103, common difference (d) = (107 – 103) = (111 – 107) = 4. And the last term (l) = 999 because the common difference remains constant.

So the number of terms in this series is calculated by the formula
$\Rightarrow {a_n} = {a_1} + \left( {n - 1} \right)d$ Where (an) is the last term of the A.P

So substitute the values in the above equation we have,
$\Rightarrow 999 = 103 + \left( {n - 1} \right)4$

Now simplify the above equation we have,
$\Rightarrow 896 = \left( {n - 1} \right)4$
$\Rightarrow n = \dfrac{{896}}{4} + 1 = 224 + 1 = 225$

So the number of terms in the series when divided by 4 leaves the remainder 3 are 225.

Now as we know 225 is the odd number
So the middle term number of this series is $\left( {\dfrac{{225 + 1}}{2}} \right) = {113^{th}}$

Now we have to calculate the sum of the series on both sides of the middle term separately.
So first calculate the value of ${112^{th}}$, ${113^{th}}$and ${114^{th}}$ terms
$\Rightarrow {a_n} = {a_1} + \left( {n - 1} \right)d$
$\Rightarrow {a_{112}} = 103 + \left( {112 - 1} \right)4$
$\Rightarrow {a_{112}} = 547$

And
$\Rightarrow {a_n} = {a_1} + \left( {n - 1} \right)d$
$\Rightarrow {a_{113}} = 103 + \left( {113 - 1} \right)4$
$\Rightarrow {a_{113}} = 551$
So the middle term of the series is 551.

And
$\Rightarrow {a_n} = {a_1} + \left( {n - 1} \right)d$
$\Rightarrow {a_{114}} = 103 + \left( {114 - 1} \right)4$
$\Rightarrow {a_{114}} = 555$

So the first series before the middle term is
103 + 107 + 111 + ....................... + 547

With first term (a1) = 103, common difference (d) = (107 – 103) = (111 – 107) = 4

And the second series after the middle term is
555 + 559 + ........................... + 999

With first term (a1) = 555, common difference (d) = (559 – 555) = 4
And the number of terms in both the series is 112.

Now as we know that the sum (Sn) of an A.P is given as
$\Rightarrow {S_n} = \dfrac{n}{2}\left( {2{a_1} + \left( {n - 1} \right)d} \right)$

Therefore sum of first series before the middle term is,
$\Rightarrow {S_{112}} = \dfrac{{112}}{2}\left( {2 \times 103 + \left( {112 - 1} \right)4} \right)$

Now simplify it we get,
$\Rightarrow {S_{112}} = 112\left( {103 + \left( {112 - 1} \right)2} \right) = 36400$

And the sum of second series after the middle term is,
$\Rightarrow {S_{112}} = \dfrac{{112}}{2}\left( {2 \times 555 + \left( {112 - 1} \right)4} \right)$

Now simplify it we get,
$\Rightarrow {S_{112}} = 112\left( {555 + \left( {112 - 1} \right)2} \right) = 87024$

So the middle term of the sequence is 551 and the sum of series before the middle term is 36400 and the sum of the second series after the middle term is 87024.

So this is the required answer.

Note – A series is said to be arithmetic progression if and only if the common difference that is the difference between the consecutive terms remains constant throughout the series. Another series which is frequently used is that of G.P in this the common ratio remains constant throughout the series. It is advised to remember basic formulas for series like A.P and G.P as it helps save a lot of time.