Question

if Sm=Sn for an AP prove that Sm+n=0.​

Answer 1

$\huge\star\frak{\underline{AnSwer:-}}$

$\underline{\bigstar\:\textsf{Given \: in \: question:}}$

$\star$ $\normalsize\sf \green{S_m \: = \: S_n}$

$\underline{\bigstar\:\textsf{ We \: have \: to \: prove }}$

$\star$ $\normalsize\sf \red{S_{m+n} \: = \: 0}$

$\underline{\bigstar\:\textsf{According \: to \: question:-}}$

$\star$ $\normalsize\textsf{\underline{Formula \: used(Sum \: of \: n \: terms)-}}$

$\star$${\underline{\boxed{\sf\green{\blue{S_n} \: = \: \orange{\frac{n}{2} } \:[\purple{2a} + \pink{(n-1)}\red{d} ]}}}}$

$\normalsize\star\textsf{\underline{Here, \: these \: values \: represents;}}$

$\normalsize\star\:\sf \blue{S_n \: = \: Sum \: of \: n \: terms}$

$\normalsize\star\:\sf \orange{n \: = \: Number \: of \: terms}$

$\normalsize\star\:\sf \purple{a \: = \: First \: term}$

$\normalsize\star\:\sf \red{d \: = \: common \: difference}$

$\underline{\bigstar\:\textsf{Let's \: Solve-}}$

$\normalsize\sf\hookrightarrow\ S_m \: = \: S_n \\ \\ \normalsize\sf\hookrightarrow\frac{m}{2} \: [ 2a +(m-1)d] \: = \: \frac{n}{2}[2a+(n-1)d] \\ \\ \normalsize\sf\hookrightarrow\frac{m}{\red{\cancel{2} } }[2a+(m-1)d \: = \: \frac{n}{\red{\cancel{2} } }[2a+(n-1)d] \\ \\ \normalsize\sf\hookrightarrow\ m[2a+(m-1)d \: = \: n[2a+(n-1)d] \\ \\ \normalsize\sf\hookrightarrow\ m[2a + dm - d] \: = \: n[2a + dn - d ] \\ \\ \normalsize\sf\hookrightarrow\ 2am + dm^2 - dm \: = \: 2an + dn^2 -dn \\ \\ \normalsize\sf\hookrightarrow\ 2am - 2an \: = \: dn^2 - dn - dm^2 + dm \\ \\ \normalsize\sf\hookrightarrow\ 2a(m-n) \: = \: d[n^2 - n - m^2 + m ] \\ \\ \normalsize\sf\hookrightarrow\ 2a(m-n) \: = \: d[(n^2 - m ^2) + (m - n)]$

$\scriptsize\sf{\: \: \: \: \: \:[ \therefore\ \: \pink{ Using \: identity: a^2-b^2 =(a+b)(a-b) }] }$

$\normalsize\sf\hookrightarrow\ 2a(m-n) \: = \: d[(n-m)(n+m) + (m-n)] \\ \\ \normalsize\sf\hookrightarrow\ 2a(m-n) \: = \: d[(-(m-n))(n+m) +(m-n)] \\ \\ \normalsize\sf\hookrightarrow\ 2a(m-n) \: = \: d(m-n)[-(m+n)+1] \\ \\ \normalsize\sf\hookrightarrow\ 2a \red{\cancel{(m-n)} } \; = \:d \red{\cancel{(m-n)} }[-(m+n)+1] \\ \\ \normalsize\sf\hookrightarrow\ 2a \: = \: d[-m-n -1]$

$\rule{300}{2}$

$\normalsize\sf\hookrightarrow\ S_{m+n} \: = \: \frac{m+n}{2}[2a +(m+n-1)d]$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \pink{ put \: the \: value \: of \: 2a}) }$

$\normalsize\sf\hookrightarrow\ S_{m+n} \: = \: \frac{m+n}{2}[d(-m-n+1) + (m+n-1)d ]$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \pink{ Cancel \: the \: values} ) }$

$\normalsize\sf\hookrightarrow\ S_{m+n} \: = \: \frac{m+n}{2}[0] \\ \\ \normalsize\sf\hookrightarrow\ S_{m+n} \: = \: 0 \\ \\ \normalsize\sf\hookrightarrow\ Hence \: prove$

Answer 2

Here is the Answer,

Sm = m/2 [ 2a + ( m -1 )d]

Sn = n/2 [2a + ( n - 1 )d]

As per question,

Sm = Sn

⇒m/2 [2a + (m-1)d] = n/2 [2a + (n-1)d]

⇒m/2 [2a + (m-1)d] - n/2 [2a + (n-1)d] = 0

⇒m [2a + md -d] -n [2a + nd - d] = 0

⇒2am + m²d - md -2an -n²d + nd = 0

⇒2am -2an + m²d - n²d - md + nd = 0

⇒2am - 2an +(m²-n²-m+n)d = 0

⇒2a(m-n) + [(m-n)(m+n - 1)]d = 0

⇒(m-n)[2a + (m+n-1)d} = 0

⇒2a + (m+n-1)d = 0 -------(1)

Now,

Sm+n = (m+n)/2 [ 2a +(m+n - 1)d]

⇒Sm+n = (m+n)/2 × 0   {From(1)}

⇒Sm+n = 0  

∵Proved

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