**CONCEPT: **-Sequences

-In an A.P, Last term (L)=(n/2)*[2a+(n-1)d] where

a=first term;

d=common difference between the terms;

n=number of terms

**SOLUTION:**

Value after 24 years=400+800+1200+.....14400

L=(n/2)[(2a+(n-1)d]

=>144000 = 24/2 [2a + (24-1)*400]

=>a=1400

Hence the value of the certificates in the 13th year = 144+ (13-1)*400

= $ 6200 (C)