12/2 (2a+11d) = 168
a+6d = 7
Now just solve for a and d.
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
3 answers
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
check my my YouTube
a = 3
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
check my my YouTube
a = 3
What's the name of your YouTube channel
1 answer