Suppose that $n=20$ , and the difference between each term is $x$ .

$a=n-2x$

$b=n-x$

$c=n+x$

$d=n+2x$

By summing all the 4 equations above, we get:

$a+b+c+d=(n-2x)+(n-x)+(n+x)+(n+2x)$

$a+b+c+d=4n$

By substituting $n=20$ , we get:

$\boxed{a+b+c+d=80}$

since a, b, 20, c, d are in AP,

20-b = c-20 (terms have same common difference between them)

that gives us, b+c=40...............(i)

similarly, b-a=d-c

that gives us b+c = a+d

so, a+b+c+d = 40+40 = 80

$Given\quad that\quad { a }_{ 1 }\quad is\quad the\quad first\quad term\quad in\quad the\quad sequence\quad and\quad d\quad be\quad \\ the\quad constant\quad difference\quad between\quad two\quad consecutive\quad terms,\quad \\ { n }^{ th }\quad element\quad can\quad be\quad defined\quad as,\qquad \qquad \\ \qquad \qquad \qquad \qquad { a }_{ n }={ a }_{ 1 }+(n-1)d\\ \qquad \\ For\quad n=3,\quad a_{ n }=20,\quad { a }_{ 1 }=a\\ \qquad 20={ a }+(3-1)d\\ Also,\quad d=b-a=20-b=c-20=d-c\\ \qquad So,\quad a=20-2(b-a)\\ \quad \\ \qquad \qquad \qquad \qquad \boxed { a=\frac { (20-2b) }{ 3 } } \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \\ The\quad sum\quad of\quad the\quad members\quad of\quad a\quad finite\quad arithmetic\quad \\ progression\quad is\quad called\quad an\quad arithmetic\quad series.\\ for\quad n\quad members,\quad the\quad sum\quad would\quad be\\ \qquad \qquad \qquad \qquad { a }_{ 1 }+...+{ a }_{ n }=\frac { n({ a }_{ 1 }+{ a }_{ n }) }{ 2 } \\ \\ So,\quad for\quad n=3,\\ \qquad \qquad \qquad \qquad a+b+20=\frac { 3(a+20) }{ 2 } \\ \\ \qquad \qquad \qquad \qquad \boxed { a=(2b-20) } \\ \\ So,\quad \frac { (20-2b) }{ 3 } =(2b-20)\\ \\ b=10\\ \\ thus,\quad our\quad AP\quad is\quad a,\quad 10,\quad 20,\quad c,\quad d\\ \\ So,\quad d=(20-10)=10\\ Thus,\quad the\quad sequence\quad is\quad 0,\quad 10,\quad 20,\quad 30,\quad 40\\ \\ and\quad a+b+c+d=80.\quad \\ \qquad$

$\text{Let the difference between each number be}$ $n$ . $\text{Then we have}$

$a=20-2n,$ $b=20-n,$ $c=20+n,$ $d=20+2n.$

$\text{The required answer is simply }$

$a+b+c+d=(20-2n)+(20-n)+(20+n)+(20+2n)=\boxed{80}.$

Yo guys,

as given a,b,20,c,d,

since it is an arithmetic progression,use common difference method....

d - c = c -20 = 20 - b = b-a,

(c - 20 = d - c) , (20 - b = c -20) , (b -a = 20 -b) , (b - a = d - c)

2c = d + 20 , b + c = 40 , 2b = a + 20 , b + c = a + d ,

noted that b + c = a + d = 40,

therefore, b + c = 40 , a + d = 40,

a + b + c + d = 40 + 40 = 80,

thanks...

$\frac{(20 \times 2) \times 5} {2}-20=\boxed{80}$

The numbers are in arithmetic progression so

$a, b, 20, c, d$ can be expressed as

$20-2k , 20-k, 20, 20+K, 20+2k$

$a+b+c+d=(20-2k)+(20-k)+(20+K)+(20+2k)=80$

Truly very simple!

$\begin{aligned} & \text{Let the arithmetic progression is }a,(a+b),(a+2b),(a+3b),(a+4b) \\ & a+(a+b)+(a+2b)+(a+3b)+(a+4b)=5a+10b=5(a+2b) \\ & it\text{ is equal with }a+b+20+c+d=5(a+2b) \\ & We\text{ know that the third term of the arithmetic progression is 20}=(a+2b) \\ & so,a+b+20+c+d=5(20) \\ & a+b+c+d=80 \\ \end{aligned}$

To spare the agony, I just assumed it began at 0 = a and it went up 10 with each number. That would make it 0+10+30+40 Nice and easy

Since the problem is ambiguous in the first place, you can take any arithmetic sequence with 20 as the 3rd term. I tried using 10 , 15, 20, 25, 30. :)

Let the common difference be (x) Then series will be ,b. ,20. , c. , d a , a+x, a+2x, a+3x, a+4x

Now a+2x=20 a=20-2x

Hence b= a+x= 20-2x+x = 20-x Similarly c= 20+ x d=20+2x

a+b+c+d = 20-2x+20-x+20+x+20+2x =80

(18 + 22)+(19+21)

Since this is a ap, the mean of all the digits will be the number in the middle, do you just need to multiply it by 4 and get the solution. The other explainations are completely apt but time consuming

Let the difference between any two successive numbers in the sequence be y. Hence, b= a+y , 20= a+2y , c=a+3y and d=a+4y . So, a+b+c+d = 4a+8y or a+b+c+d= 4(a+2y)=4*20=80 .

A,b,20,c,d I guessed that A=0 B=10 20 C=30 D=40

Therefor a+b+c+d = 0+10+30+40 =80

Yay I got it correct.

I made them all 20

a=18,b=19,c=21,d=22 Add them, 80.

Let d be the common difference.

a=20-2n b=20-n c=20+n d=20+2n a+b+c+d=(20-2n)+(20-n)+(20+n)+(20+2n) =20+20+20+20-2n-n+n+2n =80

Using arithmetic mean, (a+20)/2 = b; (b+c)/2=20 implies, ((a+20)/2 + c)/2)=20 and b+c=40; (20+d)/2 = c implies, (a+20+20+d)/4 = 20 implies, a+d=40. Thus a+b+c+d= (b+c) + (a+d) = 40+40=80

a+b+20+c+d = Sum .. Also ..as there are 5 terms and 20 being the middle term..it is the average of the all the numbers..Hence total sum must be 5*20 = 100 and thus a+b+c+d = 100-20 = 80

Using the AP properties we have:

The 1st term is a $\Rightarrow$ in this promblem is a .

The 2nd term is $a+d$ $\Rightarrow$ in this promblem is b .

The 3rd term is $a+2d$ $\Rightarrow$ in this promblem is 20

The 4th term is $a+3d$ $\Rightarrow$ in this promblem is c .

The 5th term is $a+4d$ $\Rightarrow$ in this promblem is d .

If we make $a+b+c+d$ , we get $4a+8d$ .

$4a+8d$ is 4 times more higher than the 3nd term $a+2d$ regardless of the progression.

Therefore the answer is 80.