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A $3$ digit palindrome is a $3$ digit number (not starting with zero) which reads the same backwards as forwards. for example $171$. The sum of all even $3$ digit palindromes is
(A) $22380$
(B) $25700$
(C) $22000$
(D) $22400$

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Last updated date: 13th Jun 2024
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Answer
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Hint: We should have the knowledge of what palindrome means to solve this problem. Palindrome signifies a sequence that reads the same backwards as forwards. such as $171$. Here, the sum of all even 3 digit palindromes is required, that means starting and ending with $2,4,6,8$. First of all find the series of even 3digits no. first & assume one of them as nth term . Then finding the common difference between them, apply the concept & formula of arithmetic series to get the value of n & then sum of all 3 digit even no.s of that series.

Complete step-by-step answer:
Even $3$ digit palindrome can be -
$202,212,222,232,........292,.....(i)$
$404,414,424,...............494,..(ii)$
$606,616,626,....................696,...(iii)$
$808,818,828,..................898,...(iv)$
now, from equation $(i)$
The last term is denoted as the ${\operatorname{n} ^{th}}$ term and the difference between the 2 numbers of the series is known as common difference. So, for the first equation
Let $292$ be the ${\operatorname{n} ^{th}}$term of the series with ${1^{st}}$term, $\operatorname{a} = 202$, and common difference, $\operatorname{d} = 212 - 202 = 10$
$\operatorname{tn} = a + \left( {n - 1} \right)d...(v)$
In the above ${t_n}$denotes the last term of the series that is $292$and the first term is denoted by $a$ which is $202$ and d is the common difference in the entire series in this specific case it comes out to be $10$
So, on simplifying equation $(v)$, we can calculate the value of $n$
$\therefore 292 = 202 + \left( {n - 1} \right) \times 10$
On simplifying the above equation, we get
$ \Rightarrow n = 9 + 1 = 10$
Taking the term which contain $n$on one side and other on different side we get
$\therefore $
Now, On simplifying for the value of $n$
$
   \Rightarrow 10n = 100 \\
   \Rightarrow n = \dfrac{{100}}{{10}} = 10 \\
 $
We know that, sum of arithmetic series,
$s = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] = \dfrac{n}{2}\left[ {a + 2} \right]$
Putting the values of a, n, d in this formula,
Now, we know the value of $n$ and the value of the first term as well as the common difference. Now we can calculate the sum of the arithmetic series.
$\therefore $sum of all even 3 digit palindromes $ = \dfrac{{10}}{2}\left[ {202 + 292} \right] + \dfrac{{10}}{2}\left[ {404 + 494} \right] + \dfrac{{10}}{2}\left[ {606 + 696} \right] + \dfrac{{10}}{2}\left[ {808 + 898} \right] = 22000$

Hence option (C) is the correct answer.

Note: In this type of problem, firstly, it should be recognised whether the series is in arithmetic, geometric or harmonic progression. Then the formula for respective series should be applied properly and then proceed. Concept of arithmetic series of finding nth term or sum of series should be crystal clear. Meaning of palindrome must be remembered otherwise one can’t even understand what they asked for.