(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 a

2 thoughts on “(ii) Find four consecutive terms in an A.P. whose sum is 36 and the product of the 2nd<br />and the 4th is 105. The terms are in a”

1. [tex]\huge{\underline{\boxed{\red{\mathcal{ANSWER}}}}}[/tex]

   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!!!

   Reply

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

   Reply