If an A.P. is such that find
.
Given:
To find:
We know, an = a + (n – 1)d where a is a first term or a1 and d is the common difference and n is any natural number
When n = 4:
∴ a4 = a + (4 – 1)d
⇒ a4 = a + 3d
When n = 6:
∴ a6 = a + (6 – 1)d
⇒ a6 = a + 5d
When n = 7:
∴ a7 = a + (7 – 1)d
⇒ a7 = a + 6d
When n = 8:
∴ a8 = a + (8 – 1)d
⇒ a8 = a + 7d
According to the question:
⇒ 3(a + 3d) = 2(a + 6d)
⇒ 3a + 9d = 2a + 12d
⇒ 3a – 2a = 12d – 9d
⇒ a = 3d
Now,
Hence, the value of