Answer:
\[{{1}^{st}}\]A.P. is 63, 65, 67,... \[a=63,\] \[d=65-63=2\] \[{{a}_{n}}=a+(n-1)d\] \[=63+(n-1)2\] \[=63+2n-2=61+2n\] \[{{2}^{nd}}\]A.P. is 3, 10, 17... \[a=3,\] \[d=10-3=7\] \[{{a}_{n}}=a+(n-1)d\] \[=3+(n-1)7\] \[=3+7n-7\] \[=7n-4\] According to question, \[61+2n=7n-4\] \[61+4=7n-2n\] \[65=5n\] \[n=\frac{65}{5}=13\] \[n=13\] Hence, \[{{13}^{th}}\]term of both A.P. is equal
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