Question

In an ap the sum of 4 term is 10 the product of 1 and 4 is 10 times the product of 2 and 3 term find the terms

Answer 1

Answer:

Solution:

Let the four terms of the AP be ;

a , (a + d) , (a + 2d) , (a + 3d).

where, a is the first term and d is the common difference.

Now, it is given that ;

The sum of four terms of the AP is 10.

=> a + (a + d) + (a + 2d) + (a + 3d) = 10

=> 4a + 6d = 10

=> 2( 2a + 3d ) = 10

=> 2a + 3d = 5

=> 2a = 5 - 3d. ---------(1)

Also, it is given that ,

The product of 1st and 4th terms is ten times the product of 2nd and 3rd terms.

=> a(a + 3d) = (a + d)(a + 2d)

Multiplying by 4 both sides,

=> 2×2[a(a + 3d)] = 2×2[(a + d)(a + 2d)]

=> 2a(2a + 6d) = (2a + 2d)(2a + 4d)

----------------(2)

Now, put 2a = 5 - 3d in eq-(2)

=> (5-3d)(5-3d+6d)=(5-3d+2d)(5-3d+4d)

=> (5-3d)(5+3d) = (5-d)(5+d)

=> 5^2 - (3d)^2 = 5^2 - d^2

=> 25 - 9d^2 = 25 - d^2

=> 8d^2 = 0

=> d = 0

Now , putting d = 0 in eq-(1) , we get;

=> 2a = 5 - 3d

=> 2a = 5 - 3•0

=> a = 5/2

Thus , the four terms of the AP is :

5/2 , 5/2 , 5/2 , 5/2.

Answer 2

$\huge\bigstar\underline\mathfrak\pink{Correct\:Question}$

In an AP, the sum of first four terms is 10, the product of 1st and 4th term is ten times the product of 2nd and 3rd term. Find the terms.

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$\huge\bigstar\underline\mathfrak\pink{Answer}$

The four terms are : 5/2, 5/2, 5/2 and 5/2 as the common difference is 0.

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$\huge\bigstar\underline\mathfrak\pink{Explanation}$

Let we suppose,

First term of AP = a

Second term of AP = a + d

Third term of AP = a + 2d

Fourth term of AP = a + 3d

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According to question,

$a_1$ + $a_2$ + $a_3$ + $a_4$ = 10

a + a + d + a + 2d + a + 3d = 10

4a + 6d = 10

After taking 2 as common,

2a + 3d = 5

=> 2a = 5 - 3d

=> a = $\frac{5 - 3d}{2}$

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Also,

The product of 1st and 4th term is ten times the product of 2nd and 3rd term. [ Given ]

a(a+3d) = 10[(a+d)(a+2d)]

a² + 3ad = 10[ a² + 2ad + ad + 2d² ]

a² + 3ad = 10[ a² + 3ad + 2d²]

put a = $\frac{5 - 3d}{2}$ in above equation.

We get,

$(\frac{5 - 3d}{2})^{2} + 3 \: (\frac{5 - 3d}{2})d = 10 \: {(\frac{5 - 3d}{2})}^{2} + 3 \times (\frac{5 - 3d}{2}) \times d + 2{d}^{2}$

After solving this,

we get d = 0

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Now, put the value of d in eq (i)

a =

$\frac{5 - {3 \times 0}}{2}$

a = 5/2

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Now,

First term of AP = a = 5/2

Second term of AP = a + d = 5/2 + 0 = 5/2

Third term of AP = a + 2d = 5/2 + 2(0) = 5/2

Fourth term of AP = a + 3d = 5/2 + 3(0) = 5/2

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_________________________

Hence,

The terms are : 5/2, 5/2, 5/2 and 5/2