Question

HOIT?
(A student taking a test consisting of 10 questions is told that
each questions after the first is worth 2 marks more than the
preceding question. If the third question of the test is worth 5 marks.
What is the maximum score that the student can obtain by attempting
8 questions?)
ANSWERS​

Answer 1

Answer:

given, a3= 5, d= 2, n= 8.

a3 = a+2d 

.: 5 = a + 2*2 

.: a = 5-4 = 1

S8= 8/2 [ 2*1 + (8-1)2] = 4

[ 2+ 14 ] = 64.

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Answer 2

Answer:

The maximum score that the student can obtain by attempting

8 questions is found to be 64.

Step-by-step explanation:

As each question after the first is worth 2 marks more than the preceding, the given situation gives us an arithmetic progression of common difference of 2.

Also, we are given that the third question is worth 5 marks, so we can say:

$a_3=5$

We know the expression for the nth term of an AP: $a_n=a+(n-1)d$, applying this in $a_3$, we get:

$a+(3-1)2=5$

Simplifying it, we get:

$a=1$

Now, we are asked to find the total maximum marks that can be obtained by attempting 8 questions, i.e., the sum of first 8 terms of the resulting AP:

$S_8=\frac{8}{2}(2a+7d)$

Substituting the information, we get:

$S_8=4((2\times1)+(7\times2))$

Simplifying it, we get:

$S_8=4\times16$

or we can say:

$S_8=64$

Thus, the maximum score that can be obtained by attempting 8 questions is found to be 64.