Asked by isaac

the sum of the first 12 terms of an AP is 168.if the third term is 7,find the value of common difference and first term
Answered by Steve
12/2 (2a+11d) = 168
a+6d = 7

Now just solve for a and d.

Answered by igweka Emmanuel
Sn = n/2(2a+ (n-1)d )=168

S12 = 12/2( 2a + (12-1)d )=168
=> 6(2+11d)=168
=> 12a+ 66d = 168 ...................equ (1)

Tn =a + (n-1)d
T3= a + (3-1)d ..............equ(2)


Then
Work equ(1) and 2

12a +66d=168
a + 2d = 7

a = 7-2d

Substitute "a"

12(7-2d) + 66d = 168

84 -24d+66d=168

84 +42d=168
42d = 168-84

42d = 84
d = 84/42
d =2






The sub for d = 2

We have


a + 2d=7
a + 2(2)=7

a + 4=7
a = 7-4



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a = 3


Answered by Anonymous
What's the name of your YouTube channel
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