Question

In an A.P. the third term is four times the first term, and the sixth term is 17; find the series

Answer 1

The series is 2, 5, 8, 11, 14, 17, ...

Answer 2

Let the first term of the Arithmetic Progression (A.P.) be $a$ and the common difference be $d$. The $n$-th term of an A.P. is given by the formula $a_n = a + (n-1)d$.

According to the problem statement:

1. The third term is four times the first term. $a_3 = 4 \times a_1$ $a + (3-1)d = 4a$ $a + 2d = 4a$ $2d = 4a - a$ $2d = 3a$ (Equation 1)

2. The sixth term is 17. $a_6 = 17$ $a + (6-1)d = 17$ $a + 5d = 17$ (Equation 2)

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

1. $3a - 2d = 0$
2. $a + 5d = 17$

From Equation 1, we can express $a$ in terms of $d$: $3a = 2d \implies a = \frac{2d}{3}$

Substitute this expression for $a$ into Equation 2: $\left(\frac{2d}{3}\right) + 5d = 17$

To eliminate the fraction, multiply the entire equation by 3: $3 \times \left(\frac{2d}{3}\right) + 3 \times 5d = 3 \times 17$ $2d + 15d = 51$ $17d = 51$ $d = \frac{51}{17}$ $d = 3$

Now substitute the value of $d$ back into the expression for $a$: $a = \frac{2d}{3} = \frac{2 \times 3}{3} = \frac{6}{3} = 2$

So, the first term is $a=2$ and the common difference is $d=3$.

The series is given by listing the terms: $a, a+d, a+2d, a+3d, \dots$

First term: $a_1 = 2$ Second term: $a_2 = a+d = 2+3 = 5$ Third term: $a_3 = a+2d = 2+2(3) = 2+6 = 8$ Fourth term: $a_4 = a+3d = 2+3(3) = 2+9 = 11$

And so on. The series is $2, 5, 8, 11, \dots$

Let's verify the conditions:

• Third term ($a_3$) = 8. Four times the first term ($4 \times a_1$) = $4 \times 2 = 8$. The condition $a_3 = 4a_1$ is satisfied.
• Sixth term ($a_6$) = $a+5d = 2+5(3) = 2+15 = 17$. The condition $a_6 = 17$ is satisfied.

The calculated values of $a$ and $d$ are correct. Therefore, the series is $2, 5, 8, 11, 14, 17, \dots$