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

1. $502$
2. $504$
3. $506$
4. $500$
edited | 1.4k views
+3
For maximum sum we have to go from 44 to 0 i.e total 22 element

apply formula sum of 'n' even number  = n(n+1) = 22*23 = 506
0
Simply use (n/2)(a+d) Here a =0, d = 2 and n = 23.

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
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.

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

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
= 2 ( 22 + 21 +20+19.................0)
= 2 ( 22 (22+1)/2 )
=506

option c
0
2+4+6+......44

N=22         d=2        a=2

Sum=N/2{2a+(N-1)d}

Sum=22/2{4+(21)2}

Sum=11{4+42}

Sum=11*46

Sum=506

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