The sixth term of an arithmetic progression is 37, and the sum of the first six terms is 147. We need to find the first term and the sum of the first fifteen terms.

1. Problem Description

The sixth term of an arithmetic progression is 37, and the sum of the first six terms is

1

4

7. We need to find the first term and the sum of the first fifteen terms.

2. Solution Steps

Let $a$ be the first term and $d$ be the common difference of the arithmetic progression. The $n$th term of an arithmetic progression is given by:

$a_n = a + (n-1)d$

The sum of the first $n$ terms of an arithmetic progression is given by:

$S_n = \frac{n}{2}[2a + (n-1)d]$

We are given that the sixth term is 37, so:

$a_6 = a + (6-1)d = a + 5d = 37$ (1)

We are also given that the sum of the first six terms is 147, so:

$S_6 = \frac{6}{2}[2a + (6-1)d] = 3(2a + 5d) = 147$

$2a + 5d = \frac{147}{3} = 49$ (2)

Now we have a system of two equations with two variables:

$a + 5d = 37$ (1)

$2a + 5d = 49$ (2)

Subtract equation (1) from equation (2):

$(2a + 5d) - (a + 5d) = 49 - 37$

$a = 12$

Substitute $a = 12$ into equation (1):

$12 + 5d = 37$

$5d = 37 - 12 = 25$

$d = \frac{25}{5} = 5$

So, the first term $a = 12$ and the common difference $d = 5$.

We need to find the sum of the first fifteen terms:

$S_{15} = \frac{15}{2}[2a + (15-1)d] = \frac{15}{2}[2(12) + 14(5)]$

$S_{15} = \frac{15}{2}[24 + 70] = \frac{15}{2}[94] = 15(47) = 705$

3. Final Answer

(a) The first term is

1

2. (b) The sum of the first fifteen terms is

7

0

5.