Question

The sum of first q terms of an AP is 162. The ratio of its 6th term to its 13th term is 1:2. Find the first and the 15th term of the AP.

Answer 1

Answer:

[648/q(q + 3)], [5184/q(q + 3)]

Step-by-step explanation:

Note:

(a) Sum of n terms of an AP s(n) = (n/2)[2a + (n - 1) * d]

(b) nth term of AP (an) = a + (n - 1) * d

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(i)

Sum of first q terms of an AP = 162.

⇒ 162 = (q/2)[2a + (q - 1) * d]

⇒ 324 = q[2a + (q - 1) * d]

(ii)

Given that ratio of its 6th term to its 13th term is 1:2

⇒ a₆/a₁₃ = 1/2

⇒ [(a + 5d)/a + 12d] = 1/2

⇒ 2(a + 5d) = a + 12d

⇒ 2a + 10d = a + 12d

⇒ a = 2d.

Substitute a = 2d in (i), we get

⇒ 324 = q[2(2d) + (q - 1) * d]

⇒ 324 = q[4d + (q - 1) * d]

⇒ 324 = q(4 + q - 1)d

⇒ 324 = q(3 + q)d

⇒ d = 324/q(3 + q)

First term:

⇒ a₁ = a + (n - 1) * d

       = 2d

       = 2[324/q(a + q)]

      = [648/q(3 + q)]

15th term:

⇒ a₁₅ = a + 14d

         = 2d + 16d

         = 16d

         = 16[324/q(3 + q)]

         = [5184/q(q + 3)]

Hope it helps!

Answer 2

sum of first q terms of an A.P. is 162

let first term be  'a'

common difference is 'd'

a+5d/a+12d =1/2

2a +10d = a + 12d

a=2d ------> 1

sum q = q/2(2a+(q-1)d)

162*2= q(2a+(q-1)d)

324 = 2aq + (q^2)d-qd

substitute 1

324 = 2aq +(q^2)*2a-q*2a

324 = (q +(q^2)-q)*2a

162 = (q +(q^2)-q)*a

162 =  a(q^2)