Find the four consecutive terms in an A.P
whose sum is 46 and the product of the first and third is 56 the terms are in ascending order.
Answers
Let the consecutive four terms of the AP are,
a, a + d, a + 2d, a + 3d
Where,
a = first term,
d = common difference, d > 0
Now, sum of the terms = a + a + d + a + 2d + a + 3d = 4a + 6d
According to the question,
4a + 6d = 46
2a + 3d = 23
3d = 23 - 2a
⇒ d = ...(1)
Now, product of the first and third term = a × (a+2d)
Again according to the question,
From equation (1),
By middle term splitting,,
a, a + d, a + 2d, a + 3d
Where,
a = first term,
d = common difference, d > 0
Now, sum of the terms = a + a + d + a + 2d + a + 3d = 4a + 6d
According to the question,
4a + 6d = 46
2a + 3d = 23
3d = 23 - 2a
⇒ d = ...(1)
Now, product of the first and third term = a × (a+2d)
Again according to the question,
From equation (1),
By middle term splitting,,
By zero product property,
a = 4 or a = 42
If a = 42, d < 0 ( not possible because sequence must be increasing )
if a = 4, d =
Hence, the sequence would be,
4, 11.5, 19, 26.5
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By zero product property,
a = 4 or a = 42
If a = 42, d < 0 ( not possible because sequence must be increasing )
if a = 4, d =
Hence, the sequence would be,
4, 11.5, 19, 26.5
Answer:
Step-by-step explanation:
First term = a+ d
Second term=a+d
third term=a+2d
fourth term=a+3d
sum of four terms=46
4/2(2a+3d)=46
2a+3d=23
a=23/2-3d/2
product = 56
(23-3d)(a+2d)x1/2=56
(23-3d)(23-d)=112
Excuse me I don't know how to proceed further