Question

720 was divided among A, B, C, D, E. The sum received by them was in ascending order and in Arithmetic Progression. E received 40 more than A. How much did B receive?

Answer 1

Answer:   B received an amount of 134.

Step-by-step explanation:  Given that 720 was divided among A, B, C, D, E and the sum received by them was in ascending order and in Arithmetic Progression. Also, E received 40 more than A.

We are to find the amount received by B.

Let a(1), a(2), a(3), a(4) and a(5) represents the amount received by A, B, C, D and E respectively.

Let d denotes the common difference of the given A.P.

We know that the n-th term and sum up to n terms of an A.P. with first term a and common difference d are given by

$a(n)=a(1)+(n-1)d,\\\\S_n=\dfrac{n}{2}(2a(1)+(n-1)d).$

Then, according to the given information, we have

$a(5)-a(1)=40\\\\\Rightarrow a(1)+(5-1)d-a(1)=40\\\\\Rightarrow 4d=40\\\\\Rightarrow d=\dfrac{40}{4}\\\\\Rightarrow d=10.$

Also,

$\dfrac{5}{2}(2a(1)+(5-1)d)=720\\\\\Rightarrow 5(2a(1)+4\times10)=1440\\\\\Rightarrow 2a(1)+40=\dfrac{1440}{5}\\\\\Rightarrow 2a(1)=288-40\\\\\Rightarrow 2a(1)=248\\\\\Rightarrow a(1)=\dfrac{248}{2}\\\\\Rightarrow a(1)=124.$

Thus, the amount received by B will be

$a(2)=a(1)+(2-1)d=124+1\times10=124+10=134.$

Thus, B received an amount of 134.

Answer 2

Answer:

a=x........(first term)

l=x+40......(last term)

n=5.........(no of term)

sn=720

sn=n/2(first term+ last term)

720=5/2(x+x+40)

720*2=(2x+40)5

1440=10x+200

10x=1240

x=124

a=124

sn=n/2(2a+(n-1)d

720=5/2(2*124+(5-1)d

1440=5(248+4d)

1440=1240+20d

1440-1240=20d

200=20d

d=10

a=124,..B=124+10,c=134+10........

B=124+10=134