Question

(ii) Find four consecutive terms in an A.P. whose sum is 36 and the product of the 2nd
and the 4th is 105. The terms are in ascending order.​

Answer 1

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Given:

• Sum of 4 consecutive terms of an AP = 36

• Product of 2nd and 4th term = 105

To Find:

• The 4 numbers

Assumption:

• Let the terms be a-3d, a-d, a+d, a+3d.

Solution:

⇒ Sum of the terms = 36

⇒ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 36

⇒ 4a = 36

⇒ a = 9 ------------(1)

Now, we know that:

Product of 2nd and 3rd term = 105

⇒ (a - d)*(a + 3d) = 105

⇒ a² + 3ad - ad - 3d² = 105

⇒ (9)² + 2(9)d - 3d² = 105

⇒ 81 + 18d - 3d² = 105

⇒ 18d - 3d² = 24

⇒ 3d² - 18d + 24 = 0

⇒ d² - 6d + 8 = 0

⇒ d² - 2d - 4d + 8 = 0  

⇒ d(d - 2) - 4(d - 2) = 0

⇒ (d - 2)*(d - 4) = 0

So,

⇒ d = 2     OR     d = 4 ----------(2)

The terms when d = 2:

a - 3d = 9 - 3(2) = 9 - 6 = 3

a - d = 9 - 2 = 7

a + d = 9 + 2 = 11

a + 3d = 9 + 3(2) = 9 + 6 = 15

The terms when d = 4:

a - 3d = 9 - 3(4) = 9 - 12 = -3

a - d = 9 - 4 = 5

a + d = 9 + 4 = 13

a + 3d = 9 + 3(4) = 9 + 12 = 21

Hence, the terms are 3, 7, 11, 15 OR -3, 5, 13, 21.

Verification:

1. Sum of terms:

⇒ 3 + 7 + 11 + 15      &      -3 + 5 + 13 + 21

⇒ 36

2. Product of the 2nd and the 4th term:

⇒ 7 * 15       &      5 * 21

⇒ 105

Hence verified!!!

Answer 2

here let the first term be a

followed by a+d,a+2d and a+3d

sn= n/2 (2a+(n-1)d)

36= 4/2(2a+3d)

18=(2a+3d)

a=18-3d/2

and now, for the product

(a+d)(a+3d)= 105

(18-3d+2d)(18-3d+6d)= 105×4

(18-d)(18+3d)= 420

324+54d-18d-3d²=420

-3d²+36+324=420

d²-12d-108=-140

d²-12d-108+140=0

d²-12d+32=0

d²-8d-4d+32= 0

d(d-8)-4(d-8)=0

(d-4)(d-8)=0

either d = 4

or d = 8

when d= 8

a= 18-3×8/2= -6/2= -3

a= -3

a2= -3+8=5

a3=-3+2×8=13

a4= -3+3×8= 21

so the numbers are -3, 5, 13 and 21

And again when d= 4

a= 18-3d/2= 18-12/2= 6/2= 3

the numbers are

a=3

a2= a+d= 3+4= 7

a3= a+2d= 3+2×4= 11

a4= a+ 3d= 3+3×4= 15