If $$\{a_n\}$$ is an arithmetic sequence such that $$S_m=S_n$$, then the value of $$S_{m+n}$$ is

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Uniteasy Admin · Accepted Answer

$$S_{m+n}=\dfrac{(S_m-S_n)(m+n)}{m-n}$$

$$S_{m}-S_{n} = a_1m+\dfrac{m(m-1)d}{2}-a_1n-\dfrac{n(n-1)d}{2}$$
$$=a_1(m-n)+\dfrac{(m^2-n^2-m+n)d}{2}$$
$$\dfrac{S_{m}-S_{n}}{m-n} = a_1+\dfrac{(m+n-1)d}{2}    $$
$$\dfrac{(S_{m}-S_{n})(m+n)}{m-n}$$
$$ =a_1(m+n)+\dfrac{(m+n)(m+n-1)d}{2} $$
$$= S_{m+n}$$
Therefore,
$$S_{m+n}=\dfrac{(S_m-S_n)(m+n)}{m-n}$$
If $$S_{m}=S_{n}$$, $$S_{m+n} = 0$$

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