Question

The eight term of an A.P. is half of it second term and 11th term exceed’s one third of fourth term by 1. Find the common difference​

Answer 1

Step-by-step explanation:

Given, an A.P in which,

a8 = 1/2(a2)

a11 = 1/3(a4) + 1

We know that an = a + (n – 1)d

⇒ a8 = 1/2(a2)

a + 7d = 1/2(a + d)

2a + 14d = a + d

a + 13d = 0 …… (i)

And, a11 = 1/3(a4) + 1

a + 10d = 1/3(a + 3d) + 1

3a + 30d = a + 3d + 3

2a + 27d = 3 …… (ii)

Solving (i) and (ii), by (ii) – 2x(i) ⇒

2a + 27d – 2(a + 13d) = 3 – 0

d = 3

Putting d in (i) we get,

a + 13(3) = 0

a = -39

Thus, the 15th term a15 = -39 + 14(3) = -39 + 42 = 3