Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Answers
Step-by-step explanation:
Given third term is 16
T3 = 16
a+(n-1)d = 16
Here n = 3
a is the first term
d is the difference between two terms
a+(3-1)d = 16
a+2d = 16............ (1)
Given that the 7th term exceeds the 5th term by 12
So T 7 =T5 + 12
a+6d = a+4d+12
a+6d-a-4d-12=0
2d-12 = 0
d = 6
From eqn (1)..
a+2d= 16
a+2 (6)=16
a+ 12 = 16
a= 4
The AP are
4,4+6,4+2 (6),4+3 (6),4+4 (6).......
4,10,16,22,28,34,40..........
Let first term of AP be a
and common difference be d
Then,
⤇ a + 2d = 16
⤇ a = 16 - 2d ..............................(i)
According to the question,
⤇ a + 6d = a + 4d + 12
⤇ 6d = 4d + 12
⤇ 6d - 4d = 12
⤇ 2d = 12
⤇ d =
⤇ d = 6
putting the value of d in eq(i)
a = 16 - 2d
⤇ a = 16 - 2 × 6
⤇ a = 16 - 12
⤇ a = 4
Now,
Hence,AP will be 4,10,16,22,......