We are given an arithmetic progression (A.P.). The sixth term is 37, and the sum of the first six terms is 147. (a) We need to find the first term of the A.P. (b) We need to find the sum of the first fifteen terms of the A.P.

1. Problem Description

We are given an arithmetic progression (A.P.). The sixth term is 37, and the sum of the first six terms is

1

4

7. (a) We need to find the first term of the A.P.

(b) We need to find the sum of the first fifteen terms of the A.P.

2. Solution Steps

(a) Let $a$ be the first term and $d$ be the common difference of the A.P. The $n$th term of an A.P. is given by:

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

The sum of the first $n$ terms of an A.P. 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)

We have a system of two linear equations with two variables, $a$ and $d$:

$a + 5d = 37$ (1)

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

Subtracting equation (1) from equation (2), we get:

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

$a = 12$

Thus, the first term is $a = 12$.

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

$12 + 5d = 37$

$5d = 37 - 12 = 25$

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

(b) Now we need to find the sum of the first fifteen terms. We have $a = 12$ and $d = 5$. Using the formula for the sum of the first $n$ terms:

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

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

$S_{15} = 705$

3. Final Answer

(a) The first term is

1

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

7

0

5.