Find t12 term in an arithmetic progression having t3 = 10 and t10 = −4 .
1 answer:
Answer:
t12= -8
Step-by-step explanation:
t3
a+2d=10 (1)
t10
a+9d=-4. (2)
From (1)
a=10-2d. (3)
Sub into (3) into equ (2)
a+9d=-4
10-2d+9d=-4
10+7d=-4
7d=-4-10
7d=-14
Divide both sides by 7
d=-14/7
d= -2
10=a+2d
10=a+2(-2)
10=a-4
10+4=a
14=a
a=14
t12=a+11d
= 14+11(-2)
=14-22
= -8
t12= -8
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