Question

Solve the above question

Answer 1

Given :-

◉ An AP whose,

• Sum of second term and fourth term = 28
• Product of second term and fifth term = 216

To Find :-

◉ Five consecutive terms in an AP

Solution :-

Let the first term and common difference of the given AP be a and d respectively.

It is given that, the sum of second and fourth term is 28.

We know,

⇒ aₙ = a + (n - 1)d

⇒ a + d + a + 3d = 28

⇒ 2a + 4d = 28

⇒ a + 2d = 14

⇒ a = 14 - 2d ...(1)

Also,

Product of second term and the fifth term is 216.

⇒ (a + d)(a + 4d) = 216

⇒ (14 - 2d + d)(14 - 2d + 4d) = 216

⇒ (14 - d)(14 + 2d) = 216

⇒ 196 + 28d - 14d - 2d² = 216

⇒ 2d² - 14d + 20 = 0

⇒ 2d² - 10d - 4d + 20 = 0

⇒ 2d(d - 5) - 4(d - 5) = 0

⇒ (2d - 4)(d - 5) = 0

Hence, there are two values of d i.e., 2 , 5

Case 1

Taking d = 2

Substitute d = 2 in (1)

⇒ a = 14 - 2×2

⇒ a = 10

Five consecutive terms of AP is given in the form,

⇒ a , a + d , a + 2d , a + 3d , a + 4d

⇒ 10 , 10 + 2 , 10 + 4 , 10 + 6 , 10 + 8

⇒ 10 , 12 , 14 , 16 , 18

Case 2

Taking d = 5

Substitute d = 5 in (1),

⇒ a = 14 - 5×2

⇒ a = 4

Now, the five consecutive terms of the AP is given by,

⇒ a , a + d , a + 2d , a + 3d , a + 4d

⇒ 4 , 4 + 5 , 4 + 10 , 4 + 15 , 4 + 20

⇒ 4 , 9 , 14 , 19 , 24

Answer 2

Answer:

Let, the first term of an Ap is a and tge common difference is d .

so,

The sum of the second and fourth term of an Ap is 28 .

a2 + a4 = 28

[a + ( n - 1)d] + [ a + ( n -1)d] = 28

[a + ( 2 -1 ) d ] +[ a+ ( 4 - 1) d] = 28

[ a + d]+[ a + 3d] = 28

a + d + a + 3d = 28

2a + 4d = 28

2(a + 2d) = 28

a + 2d = 14

a = 14 -2d ...(i)

Now,

The product of second term and fifth term of the Ap is 216 .

a2 × a5 = 216

( a + d ) ( a + 4d) = 216

Now , from eq(i) putting the value of a in that ,

(14 - 2d + d)(14 - 2d + 4d) = 216

(14 - d)(14 + 2d) = 216

14(14 + 2d ) -d( 14+ 2d) =216

196 + 28d - 14d -2d² = 216

196 + 14d - 2d² = 216

-2d² + 14d = 216 - 196

-2d² + 14d = 20

-2d² +14d - 20 = 0

- ( 2d² - 14d + 20 ) = 0

2d² - 14d + 20 = 0

2d² - ( 10 + 4)d + 20 = 0

2d² - 10d - 4d + 20 = 0

2d( d - 5) -4( d - 5) = 0

( d - 5) ( 2d - 4 ) =0

so, either

d - 5 = 0 .

d = 5

or,

2d - 4 = 0

2d = 4

d = 2

so , taking the value of d = 5 in first case and putting the value in eq(i) , we get

a = 14 - 2d

a = 14 - 2 × 5

a = 14 - 10

a = 4

so, a = 4 and d = 5

We know that the Ap is in the form of

a , a + d , a + 2d , a + 3d ............

4 , 4 + 5 , 4 + 2 × 5 , 4 + 3 × 5 ......

4 , 9 , 14 , 19 .........

The consequtive term of the Ap is

4 , 9 , 14 , 19 .........

Now,

when d = 2 , putting the value in eq(i) ,

a = 14 - 2d

a = 14 - 2 ×2

a = 14 - 4

a = 10

so,

a = 10 and d = 2

Now , The Ap would be -

a , a + d , a + 2d , a + 3d ........

10 , 10 + 2 , 10 + 2 × 2 , 10 + 3 × 2 ......

10 , 12 , 14 , 16 ............

The cobsequtive term of the Ap is 10 , 12 , 14 , 16 .........