Question

Find the $7^{th}$ term from the end for $A.P.11,17,23,29,\dots ,605$

Answer 1

Given $AP$ series is $11,17,23,29,....605$

Reversed A.P is $605,599,..........11.$

Here, First term $a=605$

Common difference, $598-605=-6$

$7^{th}$ term from the end of the given A.P $=7^{th}$ term of the reversed A.P

$=605+(7-1)(-6)$ [$\because n^{th}$ term of the A.P is given by $a_n=a+(n-1)d$]

$=605-36$

$=596$