For what value of n are the nth terms of aps:63,65,67......and3,10,17.

1.	For what value of n are the nth terms of aps:63,65,67......and3,10,17....equal?
Answer» $\bf{\underline{Solution}}:-$ $\bf\dag{\underline\orange{{{For\:First\:AP}}}}$ First term (a) = 63 Common difference (d) = 65 - 63 = 2 Now, we KNOW that, $\orange{\bigstar}\:\:{\underline{\boxed{\bf\purple{a_n=a+(n-1)d}}}}$ $\rm\longrightarrow\:a_n=63+(n-1)\times{2}$ $\rm\longrightarrow\:a_n=63+2n-2$ $\rm\longrightarrow\:a_n=61+2n$ $\bf\dag{\underline\green{{{For\:second\:AP}}}}$ First term (a) = 3 Common difference (d) = 10 - 3 = 7 Now, we know that, $\blue{\bigstar}\:\:{\underline{\boxed{\bf\pink{a_n=a+(n-1)d}}}}$ $\rm\longrightarrow\:a_n=3+(n-1)\times{7}$ $\rm\longrightarrow\:a_n=3+7n-7$ $\rm\longrightarrow\:a_n=7n-4$ Then, According to the QUESTION, $\rm\:61+2n=7n-4$ $\rm\longrightarrow\:61+4=7n-2n$ $\rm\longrightarrow\:65=5n$ $\rm\longrightarrow\:\cancel\dfrac{65}{5}=n$ $\rm\longrightarrow\:13=n$ $\rm\longrightarrow\:n=13$ Hence,13th term of these APs are equal.