Write the first four terms of the AP when the first term and common difference as follows.
(a) a = -1 and d =
−1/2
(b) a = 4 and d = -3
Answers
Arithmetic progression :-
• A sequence is said to be in AP (Arithmetic Progression), if the difference between its consecutive terms are equal.
• The nth term of an AP is given as ;
T(n) = a + (n-1)•d , where a is the first term and d is the common difference.
• The common difference of an AP is given as ;
d = T(n) - T(n-1)
• If the number of terms in an AP is n ( where n is odd ) ,then there will be a single middle term.
Also, [(n+1)/2]th term will be its middle term.
• If the number of terms in an AP is n ( where n is even ) ,then there will be two middle terms.
Also, (n/2)th and (n/2 + 1)th terms will be its middle terms.
• The sum up to nth terms of an AP is given as ;
S(n) = (n/2)•[2a + (n-1)•d] where a is the first term and d is the common difference.
• The nth term of an AP is also given as ;
T(n) = S(n) - S(n-1)
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Solution :-
(a)
→ First term = a = (-1)
→ common difference = d = (-1/2)
So,
→ T₂ = a + (n - 1)d
→ T₂ = (-1) + (2 - 1)(-1/2)
→ T₂ = (-1) + (-1/2)
→ T₂ = (-3/2) = second Term of AP .
Similarly,
→ T₃ = T₂ + d or, = a + (n - 1)d
→ T₃ = (-1) + (3 - 1) * (-1/2)
→ T₃ = (-1) + 2 * (-1/2)
→ T₃ = (-1) + (-1)
→ T₃ = (-2) = Third Term of AP.
Similarly,
→ T₄ = T₃ + d
→ T₄ = (-2) + (-1/2)
→ T₄ = (-5/2) = Fourth Term of AP.
Hence, The Required AP will be :- (-1) , (-3/2) , (-2) , (-5/2).
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(b)
→ First term = a = 4
→ common difference = d = (-3)
So,
→ T₂ = a + (n - 1)d
→ T₂ = 4 + (2 - 1)(-3)
→ T₂ = 4 - 3
→ T₂ = 1 = second Term of AP .
Similarly,
→ T₃ = T₂ + d or, = a + (n - 1)d
→ T₃ = 4 + (3 - 1) * (-3)
→ T₃ = 4 + 2 * (-3)
→ T₃ = 4 + (-6)
→ T₃ = (-2) = Third Term of AP.
Similarly,
→ T₄ = T₃ + d
→ T₄ = (-2) + (-3)
→ T₄ = (-5) = Fourth Term of AP.
Hence, The Required AP will be :- 4 , 1 , (-2) , (-5).
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Answer:-
(a) Given:
a = - 1
d = - 1/2.
We know that,
nth term of an AP = a + (n - 1)d
We already have a 1 → - 1
To find 2nd term put the value of n as 2.
→ a(2) = - 1 + (2 - 1)( - 1/2)
→ a(2) = - 1 - 1/2
→ a(2) = (- 2 - 1)/2 = - 3/2
Similarly,
a(3) = - 1 + (3 - 1)(- 1/2)
→ a(3) = - 1 + (2)(- 1/2)
→ a(3) = - 1 - 1
→ a(3) = - 2
a(4) = - 1 + (4 - 1)( - 1/2)
→ a(4) = - 1 - 3/2
→ a(4) = (- 2 - 3)/2 = - 5/2.
(b) a = 4
d = - 3
a(2) = 4 + (2 - 1)( - 3)
→ a(2) = 4 - 3
→ a(2) = 1
a(3) = 4 + (3 - 1)( - 3)
→ a(3) = 4 - 6
→ a(3) = - 2
a(4) = 4 + (4 - 1)( - 3)
→ a(4) = 4 - 9
→ a(4) = - 5
Therefore,
From (a) - The first 4 terms are - 1 , - 3/2 , - 2 , - 5/2.
From (b) - The first 4 terms are 4 , 1 , - 2 , - 5.