edited by
5,601 views
17 votes
17 votes

What will be the maximum sum of $44, 42, 40, \dots$ ?

  1. $502$
  2. $504$
  3. $506$
  4. $500$
edited by

5 Answers

Best answer
22 votes
22 votes

This is in AP.

Maximum sum means we do not need to consider negative numbers and can stop at $0.$

First find number of terms using the formula $\large a{_n} = a + (n-1)d$
Here,
$a = 44,$
$d = 42-44 = -2,$
$a{_n} = 0.$

Therefore, $0 = 44 + (n-1)(-2)$
$\Rightarrow n = 23.$

Now, sum of $n$ terms of AP is given by: $S{_n} =\dfrac{n}{2}[a+a{_n}] =\dfrac{23}{2}[44+0] = 506.$
Option C is correct!

edited by
15 votes
15 votes
Sequence is

2+4+6+... + 42+ 44.

Then we can convert it into AP

2(1+2+3+... 21+22)=> Which is

2( (22*23)/2) (Using Sum of First n no , n * n+1 / 2 => 2(11*23) => C 506.
7 votes
7 votes

it is an AP series sum till nth term formula =n/2[2a+(n-1)d] where a is first term,d is difference,n is number of terms.in this case n=22;a=2;d=2 now find out the sum whis results to 506

2 votes
2 votes
It is decreasing sequence(AP) with d=-2 For max sum all terms should be positive so lets find first negative term a+(n-1)d 44+(n-1)*-2
Answer:

Related questions

17 votes
17 votes
3 answers
1
21 votes
21 votes
2 answers
2
18 votes
18 votes
6 answers
3
24 votes
24 votes
1 answer
4
Arjun asked Sep 24, 2014
7,498 views
Find the sum of the expression$\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+............+\frac{1}{\sqrt{80}+\sqrt{81}}$$7$$8$$9$$10...