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Find t12 term in an arithmetic progression having t3 = 10 and t10 = −4 .

User Tamjid
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1 Answer

3 votes

Answer:

t12= -8

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

User Keaplogik
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