Question #88190

1 Answer
Feb 6, 2017

13+(n-1)*(-5/3)13+(n1)(53)

Explanation:

I think this is an arithmetic progression. Therefore,
T_n=a+(n-1)dTn=a+(n1)d, where aa = first term, nn = number of term and dd=common different.

13+27*(-5/3) = 13+(28-1)*(-5/3)=T_2813+27(53)=13+(281)(53)=T28

Therefore we can say that,
a=13, d=(-5/3)a=13,d=(53)

so, T_n = 13+(n-1)*(-5/3)Tn=13+(n1)(53)