Class 10

Math

All topics

Arithmetic Progressions

Third term of an $A.P.$ Is $16$ and $7_{th}$ term is $12$ more than $5_{th}$ term, then find $AP$.

Given $a_{3}=16$

$a+(3−1)d=16$

$⇒a+2d=16 ...(i)$

According to question, $a_{7}=12−a_{5}$

$⇒a_{7}−a_{5}=12$

$[a+(7−1)d]−[a+(5−1)d]=12$

$⇒a_{7}+6d−a−4d=12$

$⇒a+6d−a−4d=12$

$⇒2d=12$

$d=212 =6$

Substituting value of $d$ in equation $(i)$

$a+2(6)=16$

$∴a=16−12=4$

$AP.a,a+d,a+2d,...$

$=4,4+6,4+2×6,....$

$=4,10,16,....$

Hence, required $A.P.$ is $4,10,16,22,...$