Question

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

Answer 1

Answer:

first term=a

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

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Step-by-step explanation:

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Answer 2

Sequence would be 4, 11.5, 19, 26.5

Step-by-step explanation:

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 = $\frac{23-2a}{3}$      ...(1)

Now, product of the first and third term = a × (a+2d)

Again according to the question,

$a(a+2d)=56$

From equation (1),

$a(a+\frac{46-4a}{3}) = 56$

$a(\frac{3a+46-4a}{3})=56$

$a(\frac{46-a}{3}) = 56$

$46a-a^2=168$

$\implies a^2 - 46a + 168 =0$

By middle term splitting,,

$a^2-42a-4a+168=0$

$a(a-42)-4(a-42)=0$

$(a-4)(a-42)=0$

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 = $\frac{23-2(4)}{3}=\frac{23-8}{3}=\frac{15}{3}=5$

Hence, the sequence would be,

4, 11.5, 19, 26.5

#Learn more:

Find the four consecutive terms in an A.P . whose sum is -54 and the sum of first and third term is -30.

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