Question

The third and the fifth terms of an arithmetic progression are 10 and -10 respectively.(a) determine the first term and the common diference.(b) the sum of the first 15 terms

Answer 1

Yes

Answer 2

(a)
nth term=a+(n-1)d
10=a+(3-1)d
10=a+2d
a+2d=10....eqn (i)

-10=a+(5-1)d
-10=a+4d
a+4d=-10....eqn (ii)

a+2d=10
a+4d=-10 solve by elimination method
-2d=20
d=-10
substitute d=-10 in the eqn a+2d=10
a+(2×-10)=10
a-20=10
a=30
therefore, the first term is 30 and the common difference is -10

(b)
Sn=n/2[2a+(n-1)d]
=15/2[2×30+(15-1)×-10]
=15/2[60+(14×-10)]
=15/2[60-140]
=15/2×-80
=-600