Question

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

Answer 1

Given third term of A.P is $a_3=16$

and $7$th term exceeds its $5$th term by $12$.

$⟹a_7=a_5+12$

nth term of an A.P is given by

$a_n=a_1+(n-1)d$

hence 3rd term is

$a_3=a_1+(3-1)d$

put $$a_{3}=16$ we get

$16=a_1+2d....eq\left(1\right)$

5th term of A.P is given by

$a_5=a_1+(5-1)d$

$a_5=a_1+4d$

7th term of A.P is given by

$a_7=a_1+(7-1)d$

$a_7=a_1+6d$

and given that $⟹a_7=a_5+12$

hence

$a_1+6d=a_1+4d+12$

$⟹a_1-a_1+6d-4d=12$

$⟹2d=12$

$⟹d=\mathrm{6.....}eq\left(2\right)$

put d=6 in eq(1) we get

$16=a_1+2\times 6$

$⟹16=a_1+12$

$⟹a_1=16-12$

$a_1=4$ first term of A.P

second term of A.P is

$a_2=a_1+d=4+6=10$

3rd term of A.P is

$a_3=a_2+d=10+6=16$

4th term of A.P is

$a_4=a_3+d=16+6=22$

and so on.

hence te A.P is

$4,10,16,22,...\dots .$