Q. 3 B3.7( 11 Votes )

# Divide 20 into four parts which are in A.P. such that the ratio of the product of the first and fourth is to the product of the second and third is 2 : 3.

Answer :

Let the four parts which are in AP are

(a – 3d), (a – d), (a + d), (a + 3d)

According to question,

The sum of these four parts = 20

⇒(a – 3d) + (a – d) + (a + d) + (a + 3d) = 20

⇒ 4a = 20

⇒ a = 5 …(i)

Now, it is also given that

product of the first and fourth : product of the second and third = 2 : 3

i.e. (a – 3d) × (a + 3d) : (a – d) × (a + d) = 2 : 3

[∵(a – b)(a + b) = a^{2} – b^{2} ]

⇒ 3(a^{2} – 9d^{2}) = 2(a^{2} – d^{2})

⇒ 3a^{2} – 27d^{2} = 2a^{2} – 2d^{2}

⇒ 3a^{2} – 2a^{2} = – 2d^{2} + 27d^{2}

⇒ (5)^{2} = – 2d^{2} + 27d^{2} [from (i)]

⇒ 25 = 25d^{2}

⇒ 1 = d^{2}

⇒ d = ±1

Case I: if d = 1 and a = 5

a – 3d = 5 – 3(1) = 5 – 3 = 2

a – d = 5 – 1 = 4

a + d = 5 + 1 = 6

a + 3d = 5 + 3(1) = 5 + 3 = 8

Hence, the four parts are

2, 4, 6, 8

Case II: if d = – 1 and a = 5

a – 3d = 5 – 3( – 1) = 5 + 3 = 8

a – d = 5 – ( – 1) = 5 + 1 = 6

a + d = 5 + ( – 1) = 5 – 1 = 4

a + 3d = 5 + 3( – 1) = 5 – 3 = 2

Hence, the four parts are

8, 4, 6, 2

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The sum of n terms of an A.P. is 3n^{2}+ 5n. Find the A.P. Hence, find its 16th term.