The sum of the first $m$ terms of an AP is $n$, and the sum of its first $n$ terms is $m$ (where $m
eq n$). Show that the sum of its first $(m+n)$ terms is $-(m+n)$.
Step-by-Step Solution
Key Concept: Write both sum conditions using the sum formula, subtract to find a relation, and use it to directly evaluate S(m+n).
$S_m=n\Rightarrow\dfrac m2[2a+(m-1)d]=n\Rightarrow2am+m(m-1)d=2n$ — (1). [1.0 Mark]
$S_n=m\Rightarrow\dfrac n2[2a+(n-1)d]=m\Rightarrow2an+n(n-1)d=2m$ — (2). [1.0 Mark]
Subtracting (2) from (1): $2a(m-n)+d[m(m-1)-n(n-1)]=2(n-m)$. Since $m(m-1)-n(n-1)=(m^2-n^2)-(m-n)=(m-n)(m+n-1)$: $2a(m-n)+d(m-n)(m+n-1)=-2(m-n)$. [1.5 Marks]
Dividing throughout by $(m-n)$ (valid since $m
eq n$): $2a+d(m+n-1)=-2$. [1.0 Mark]
Now, $S_{m+n}=\dfrac{m+n}{2}[2a+(m+n-1)d]=\dfrac{m+n}{2}\times(-2)=-(m+n)$. Hence proved. [0.5 Mark]
Correct Answer: