Question

which term of the AP 8, 14, 20, 26---- will be 72 more than its 41 st term?​

Answer 1

Answer:-

$\sf \huge \: AP:8,4,20,26.... \\ \: \\ \\ \sf \huge \mapsto \: a=8 \\ \\ \sf \huge \mapsto \: d=14-8=6 \\ \\ \bf \huge \: Let T_n=T_41+72 \\ \\ \tt \huge \longmapsto \: a+(n-1)d-a-40d=72 \\ \\ \tt \huge \longmapsto \:d(n-1-40)=72 \\ \\ \tt \huge \longmapsto \:6(n-41)=72 \\ \\ \tt \huge \longmapsto \:n-41=12 \\ \\\tt \huge \longmapsto \: n=12+41 \\ \\ \tt \huge \longmapsto \:n=53 \\ \\ </p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p> \bf \huge \bigstar \: Required \: \: term \: \: is \: \: 53 \\ \\ </p><p></p><p>\bf \huge \bigstar \:53 \: th \: \: term \: \: of \: \: AP \: \: will \: \: be \: 72 \: \: more \: \: than \: \: 41 \: \: th \: \: term.....</p><p>$

Answer 2

Step-by-step explanation:

$$\begin{lgathered}\sf \huge \: AP:8,4,20,26.... \\ \: \\ \\ \sf \huge \mapsto \: a=8 \\ \\ \sf \huge \mapsto \: d=14-8=6 \\ \\ \bf \huge \: Let T_n=T_41+72 \\ \\ \tt \huge \longmapsto \: a+(n-1)d-a-40d=72 \\ \\ \tt \huge \longmapsto \:d(n-1-40)=72 \\ \\ \tt \huge \longmapsto \:6(n-41)=72 \\ \\ \tt \huge \longmapsto \:n-41=12 \\ \\\tt \huge \longmapsto \: n=12+41 \\ \\ \tt \huge \longmapsto \:n=53 \\ \\ \bf \huge \bigstar \: Required \: \: term \: \: is \: \: 53 \\ \\ \bf \huge \bigstar \:53 \: th \: \: term \: \: of \: \: AP \: \: will \: \: be \: 72 \: \: more \: \: than \: \: 41 \: \: th \: \: term.....\end{lgathered}$$