AP EAPCET 25-Aug-2021 Shift 2 Paper

Given f:R→R is defined as
f(x+y)=f(x)+f(y),∀x,y‌in‌R and f(1)=5
To Find

n

∑

r=1

f(r)= ?
Since, f(x+y)=f(x)+f(y) ...(i)
Let x=y=1
f(1+1)=f(1)+f(1)=5+5
f(2)=10
Again in Eq. (i), put x=2,y=1
f(2+1)=f(2)+f(1)
f(3)=10+5
f(3)=15
Similarly, we can find f(4)=15+5=20
f(5)=25,...f(n)=5n

n

∑

r=1

f(r)=f(I)+f(2)+...+f(n)
=5+10+15+...+5n
which forms an AP of n-terms with first term (a)=5 and common difference (d)=5
=

n

2

[2(5)+(n−1)5]
=

n

2

[10+5n−5]=

n

2

[5n+5]

n

∑

r=1

f(r)=

5n(n+1)

2