sum of 2 terms of an arithmetic sequence is 32 and the sum of first 3 terms is 63. Find the 4th term of the sequence
Answers
Given :-
- Sum of two terms of an AP is 32.
- Sum of three terms of an AP is 63.
To find :-
- 4th term of AP?
Solution :-
We know that,
Sum of n terms of an AP is,
Sₙ = n/2 (2a + (n - 1)d)
★ Sum of two terms of an AP is 32.
➻ S₂ = 2/2( 2a + (2 - 1)d)
➻ 32 = 2a + d
➻ d = 32 - 2a [ eq (1) ]
★ Sum of three terms of an AP is 63.
➻ S₃ = 3/2 (2a + (3 - 1)d)
➻ 63 = 3/2 (2a + 2d)
➻ 63 × 2/3 = 2a + 2d
➻ 42 = 2a + 2d [ eq (2) ]
Now, Putting value of eq (1) in eq (2),
➻ 42 = 2a + 2(32 - 2a)
➻ 42 = 2a + 64 - 4a
➻ 42 = 64 - 2a
➻ 2a = 64 - 42
➻ 2a = 22
➻ a = 22/2
➻ a = 11
Now, Putting value of a in eq (1),
➻ d = 32 - 2 × 11
➻ d = 32 - 22
➻ a = 10
Now, We have to find 4th term,
➯ a₄ = 11 + 3 × 10
➯ a₄ = 11 + 30
➯ a₄ = 41
∴ Hence, 4th term of the sequence is 41.
Given : sum of 2 terms of an arithmetic sequence is 32 and the sum of first 3 terms is 63.
To Find : 4th term of the sequence
Solution:
Let say
AP is
a , a + d , a + 2d , a + 3d and so on
a + a + d = 32
=> 2a + d = 32
a + a + d + a + 2d = 63
=> 3a + 3d = 63
=> a + d = 21
2a + d = 32
a + d = 21
=> a = 11
=> d = 10
4th term of the sequence = a + 3d
= 11 + 3(10)
= 41
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