Question

If the series is in AP with common difference 2 & T_(1) T_(3) T_(7) are in GP find T_(2)​

Answer 1

Answer:

6

Step-by-step explanation:

$t_{1}=a\\t_{3}=a+2d=a+4\\t_{7}=a+6d=a+12\\\\t_{1}.t_{7}=t_{3}^{2}\\ a(a+12)=(a+4)^2\\a^2+12a=a^2+8a+16\\4a=16\\a=4\\t_{2}=a+d=4+2=6$