Someone help me with this ASAP

here's the solution,

let the segment be AB, and PQ is perpendicular bisector of AB at Q , and Let P be the point on perpendicular bisector, and join P with A and B,

and since PQ bisects AB into AQ and BQ, they are equal,

that is , AQ = BQ

now, two triangles are formed.

In the given triangles, PQA and PQB

( common side )

( perpendicular forms 90° )

( they are equal, proved above )

here, Triangle PQA and PQB are congruent by SAS ( side angle side ) criterion.

So, AP = BP ( by C.P.C.T )

= 5

step-by-step explanation:

to find the 15th term substitute n = 15 into the nth term formula

= (- 3 × 15 ) + 50 = - 45 + 50 = 5