Question

Solve for $x:25+22+19+......+x=115$
If a,b,c are in A.P, Prove a+b,b+c,c+a are in A.P

Answer 1

$a_0=25d=-3S=\frac{n}{2}(2a_0+(n-1\left)d\right)\Rightarrow 115=\frac{n}{2}(2\times 25+(n-1\left)(-3)\right)\Rightarrow 230=n(50+3-3n)\Rightarrow {3n}^2-53n+230\Rightarrow 3n(n-30)-23(n-10)\Rightarrow (3n-23)(n-10)\Rightarrow n=10,n=\frac{23}{3}x=a_0+(n-1)d\Rightarrow x=25+9\times -3\Rightarrow x=-2$

Let us assume that

$(a+b),(b+c),(c+a)$ are in A.P

Then

$(a+b)+(c+a)=2(b+c)\Rightarrow 2a+b+c=2b+2c\Rightarrow 2a=b+c$ is true because a,b,c are in A.P

Hence proved