please answer question no.2!!!
Answers
⇒ We know that nth term of an AP an = a + (n - 1) * d.
⇒ Sum of first n terms of an AP sn = (n/2)[2a + (n - 1) * d].
According to question,
(i)
Given that Sum of third and seventh terms of an AP is 6.
⇒ a₃ + a₇ = 6
⇒ a + (3 - 1) * d + (a + 7 - 1) * d = 6
⇒ a + 2d + a + 6d = 6
⇒ 2a + 8d = 6
⇒ a + 4d = 3
⇒ a = 3 - 4d ----- (1)
(ii)
Given that their product is 8.
⇒ (a + 2d) * (a + 6d) = 8
⇒ (3 - 4d + 2d) * (3 - 4d + 6d) = 8
⇒ (3 - 2d) * (3 + 2d) = 8
⇒ 9 - 4d^2 = 8
⇒ 4d^2 = 1
⇒ d = 1/2.
When d = 1/2:
⇒ a = 3 - 4d
= 3 - 4(1/2)
= 1.
When d = -1/2:
⇒ a = 3 - 4d
= 3 - 4(-1/2)
= 5.
Now,
We need to calculate the Sum of first sixteen terms of the AP:
When a = 1 and d = (1/2):
⇒ (16/2)[2 + (16 - 1) * (1/2)]
⇒ 8[2 + 15/2]
⇒ 4[19]
⇒ 76.
When a = 5 and d = (-1/2):
⇒ (16/2)[2(5) + (16 - 1) * (-1/2)]
⇒ 8[10 - 15/2]
⇒ 4[5]
⇒ 20.
Therefore, Sum of first sixteen terms of the AP is 76 and 20.
Hope it helps!
