The 9th term of an Arithmetic progression is 52 while the 12th term is 70. Find the product of half of first term and two-third of 5th term

Answer 1

We know that the nth term of an arithmetic progression can be found by the formula:

Term_n = a + (n-1)*d

where 'a' is the first term and 'd' is the common difference.

Given that:
Term_9 = a + 8*d = 52 -----> (equation 1)
Term_12 = a + 11*d = 70 -----> (equation 2)

We can solve these two equations to find 'a' and 'd'. Subtract equation 1 from equation 2 to get:

3*d = 18 --> d = 6

Substitute d = 6 into equation 1 to find 'a' :

a + 8*6 = 52 --> a = 52 - 48 = 4

We want to find the product of half of first term and two-thirds of fifth term. That is, (1/2)*a * (2/3)*Term_5.

Substituting a = 4 and Term_5 = a + 4*d = 4 + 4*6 = 28, we find:

Product = (1/2)*4 * (2/3)*28 = 2 * (56/3) = 112/3 = 37.33
This is the required product.