Question

In a certain test ${a_i}$ students gave wrong answers to at least $i$ questions, where $i = 1,2,...,k$.
No student gave more than $k$ wrong answers. The total no. of wrong answers given in……..
A) ${a_1} + {a_2} + {a_3} + ... + {a_k}$
B) ${a_1} + {a_2} + {a_3} + ... + {a_{\left( {k - 1} \right)}}$
C) ${a_1} + {a_2} + {a_3} + ... + {a_{\left( {k + 1} \right)}}$
D) None of the above

Answer 1

Hint:Here in this question the sequence of students who gave how many wrong answers are given.So, by this we can calculate how many students gave one wrong answer, further on we can get how many students gave two wrong answers, so keeping this on and by calculating further we will get how many students gave wrong answers. After that by multiplying each of them by the no. of wrong answers they have given and then adding them all, we will get our answer.

Complete step-by-step answer:
First of all we will start by enlisting the given information in the question.
So, it is given that a test was conducted and the no. of students who gave wrong answers to at least $i$ questions is ${a_i}$. Where, $i = 1,2,...,k$.
And no student gave more than$k$ wrong answers.
So, from the above information we can say that
The Number of students gave wrong answer to exactly one question$ = $
(No. of students gave at least one wrong answer)$ - $ (No. of students gave at least two wrong answer)
$ \Rightarrow $The Number of students gave wrong answers to exactly one question$ = {a_1} - {a_2}$.
Similarly, The Number of students gave wrong answers to exactly two questions$ = {a_2} - {a_3}$.
Further on, The Number of students gave the wrong answer to exactly three questions$ = {a_3} - {a_4}$.
And so on we will all value in this sequence.
Now, the total wrong answers will be equal to the sum of the products of the no. of wrong answers with the no. of students who gave exactly that much wrong answers.
$ \Rightarrow $Total wrong answers$ = $ $1 \times $(No. of students gave exactly one wrong answer)$ + 2 \times $(No. of students gave exactly two wrong answer)$ + 3 \times $(No. of students gave exactly three wrong answer)……………upto $k$ terms
$ \Rightarrow $ Total wrong answers$ = 1\left( {{a_1} - {a_2}} \right) + 2\left( {{a_2} - {a_3}} \right) + 3\left( {{a_3} - {a_4}} \right)... + k\left( {{a_{\left( {k - 1} \right)}} - {a_k}} \right)$
$ \Rightarrow $ Total wrong answers$ = {a_1} - {a_2} + 2{a_2} - 2{a_3} + 3{a_3} - 3{a_4}... + k{a_{\left( {k - 1} \right)}} - k{a_k}$
$ \Rightarrow $ Total wrong answers$ = {a_1} + {a_2} + {a_3}... + {a_k}$
$ \Rightarrow $Option A is correct.

Note: In these types of questions generally constructing the equation of the no. of students who gave exactly one wrong or two wrong or etc was tricky, and on that step students may make a mistake, for resolving that read the question carefully and then start attempting the question.