Question

if m times of mth term of an AP is equal to nth times of nth term then find the m+n th term of the AP​

Answer 1

$\Large\underline{\textsf{\textbf{\purple{ \mapsto Given:}}}}$

• m times the mth term of an AP is equal to nth times of nth term .

$\Large\underline{\textsf{\textbf{\purple{ \mapsto To\:Find:}}}}$

• The ( m + n )th term of the AP .

$\Large\underline{\textsf{\textbf{\purple{ \mapsto Answer:}}}}$.

Given that m times the mth term of an AP is equal to nth times of nth term . So , we know the formula to find the k th term of an Arthemetic Progression as ;

$\qquad\qquad\large\boxed{\pink{\bf\blue{\dag}\:T_k\:=\:a+(k-1)d}}$

Where ,

• $\red{\sf T_k \:is\: the\: required\:term.}$
• $\red{\sf a \:is\: the\:first\:term.}$
• $\red{\sf d\:is\: the\: common\: difference.}$

$\underline{\underline{\green{\mapsto \sf mth\:term\:of\: the\:AP:-}}}$

$\tt:\implies T_k = a + (k-1)d$

$\bf\implies T_m = a + (m-1)d$

$\rule{200}2$

$\underline{\underline{\green{\mapsto \sf nth\:term\:of\: the\:AP:-}}}$

$\tt:\implies T_k = a + (k-1)d$

$\bf\implies T_m = a + (n-1)d$

$\rule{200}2$

$\underline{\underline{\pink{\mapsto \sf \mathscr{A} ccording\:\;\mathscr{T}o\:\; \mathscr{Q} uestion\::-}}}$

$\tt:\implies m( T_m) = n( T_n)$

$\tt:\implies m[ a + (m-1)d = n[ a + (n-1)d]$

$\tt:\implies am + m(m-1)d = an + n(n-1)d$

$\tt:\implies am + m^2d - md = an + n^2d - nd$

$\tt:\implies ( am + m^2d - md ) - ( an + n^2d - nd ) = 0$

$\tt:\implies am + m^2d-md -an -n^2d+nd = 0$

$\tt:\implies am - an + m^2d - n^2d -md+nd$

$\tt:\implies a( m -n ) + d ( m^2-n^2) - d ( m - n)=0$

$\tt:\implies a(m-n) + d(m+n)(m-n)-d(m-n) = 0$

$\tt:\implies (m-n) [ a + d(m+n) - d ] = 0$

$\tt:\implies [ a + d(m+n) - d ] =\dfrac{0}{m-n}$

$\tt:\implies a + d(m+n)-d = 0$

$\tt:\implies a + \lgroup ( m + n) - 1 \rgroup d = 0$

$\underline{\boxed{\red{\tt{\longmapsto T_{m+n}=0}}}}$

$\boxed{\green{\pink{\dagger}\bf Hence\: the\:(m+n)th\:term\:is\:0.}}$

Answer 2

$\mapsto\sf \color{blue} \: \huge Given$

• M term of mth term of an AP is equal to nth times of nth.

$\sf \mapsto\color{cyan} \huge Required \: to \: Find$

The m + n th term of the AP

$\sf \mapsto \color{pink} \: \huge Solution$

$\sf \: T_{n} = a + (n - 1)d$

$\sf \: n \: T_{n} = m \: T_{n}$

$\sf \implies \: n(a + (n - 1)d) = m(a + (m - 1)d)$

$\sf \implies \: n \: a + {n}^{2} \: d \: - nd = a \: m \: + {m}^{2} \: d - md$

$\sf \implies \: a(n - m) + d( {n}^{2} - {m}^{2} ) - (n - m) \: d \: = 0$

$\sf \implies \: (n - m)(a + d(n + m - 1)) = 0$

$\sf \implies \: a + (n + m - 1) \: d = 0$

$\bf \implies \color{red} \: T_{m + n} = 0$

$\bf \color{green}Therefore \: the \: m + n \: th \: term \: of \: the \: AP \: is \: 0$