Question

Find the arithmetic progression whose third tem is 16 and seventh term exceeds its fifth term by 12.

Answer 1

$T_3$ =16

$\Rightarrow$ a+2d=16

$T_7$ = $T_5$ + 12

$T_7$ = a+4d +12...... Eq (1 )

In A.P , common difference between the terms is constant .
So, let common difference =d


Also , $T_7$ = $T_5$ +d+d

= a+ 4d +2d

= a + 6d ..........Eq (2)
fromEq 1 and Eq 2 , we get
a+ 4d+12 = a+ 6d
or 12 =2d
or d=6
therefore ,
2nd term in A.P= a+d =16+6 =22
3rd term = a+2d = 16+2*6 =16+12=28
and the 7th term =a+6d =16+6*6=16+36 =52

Hence the A.p will be 16,22,28,34,40,46,52