Determine the A.P. whose third term is 11 and seventh term exceeds the fifth term by 8.
Answers
Step-by-step explanation:
Solution :-
Let a be the 1st term and d be the common difference of the AP.
We have,
a₃ = 16
a + 2d = 16 ..... (i)
a₇ = a + 6d and a₅ = a + 4d
According to the questions,
⇒ a₇ - a₅ = 12
⇒ a + 6d - (a + 4d) = 12
⇒ a + 6d - a - 4d = 12
⇒ 2d = 12
⇒ d = 12/2
⇒ d = 6
Putting the value of d in equation (i), we get
a + 2 × 6 = 16
a + 12 = 16
=>a = 16 - 12 = 4
Here, a = 4, d = 6
Hence,
AP is a, a + d, a + 2d, a + 3d +..
4, 4 + 6, 4 + 12, 4 + 18, ...
4, 10, 16, 22..
Answer:
The required AP is 3, 7, 11, 15,..
Step-by-step-explanation:
Let the first term of AP be a and the common difference be d.
We have given that,
t₃ = 11
We know that,
tₙ = a + ( n - 1 ) * d
∴ t₃ = a + ( 3 - 1 ) * d
⇒ 11 = a + 2d
⇒ a = 11 - 2d - - - ( 1 )
From the given condition,
t₇ = t₅ + 8
⇒ a + ( 7 - 1 ) * d = a + ( 5 - 1 ) * d + 8
⇒ a + 6d = a + 4d + 8
⇒ a + 6d - a - 4d = 8
⇒ 2d = 8
⇒ d = 8 / 2
⇒ d = 4
By substituting d = 4 in equation ( 1 ), we get,
a = 11 - 2d - - - ( 1 )
⇒ a = 11 - 2 * 4
⇒ a = 11 - 8
⇒ a = 3
Now,
t₁ = a = 3
t₂ = t₁ + d = 3 + 4 = 7
t₃ = t₂ + d = 7 + 4 = 11
t₄ = t₃ + d = 11 + 4 = 15
∴ The required AP is 3, 7, 11, 15,..