# The Sum of First Seven Terms of an A.P. is 182. If Its 4th and the 17th Terms Are in the Ratio 1 : 5, Find the A.P. - Mathematics

Sum

The sum of first seven terms of an A.P. is 182. If its 4th and the 17th terms are in the ratio 1 : 5, find the A.P.

#### Solution

Let a be the first term and d be the common difference.
We know that, sum of first n terms = Sn = $\frac{n}{2}$[2a + (n − 1)d]

According to the question,

$S_7 = 182$
$\Rightarrow \frac{7}{2}\left[ 2a + \left( 7 - 1 \right)d \right] = 182$
$\Rightarrow \frac{1}{2}\left( 2a + 6d \right) = 26$
$\Rightarrow a + 3d = 26$
$\Rightarrow a = 26 - 3d . . . . (1)$

Also,

$\frac{a_4}{a_{17}} = \frac{1}{5}$

$\Rightarrow \frac{a + (4 - 1)d}{a + (17 - 1)d} = \frac{1}{5}$

$\Rightarrow \frac{a + 3d}{a + 16d} = \frac{1}{5}$

$\Rightarrow 5(a + 3d) = a + 16d$

$\Rightarrow 5a + 15d = a + 16d$

$\Rightarrow 5a - a = 16d - 15d$

$\Rightarrow 4a = d . . . . (2)$

On substituting (2) in (1), we get

$a = 26 - 3\left( 4a \right)$
$\Rightarrow a = 26 - 12a$
$\Rightarrow 12a + a = 26$
$\Rightarrow 13a = 26$
$\Rightarrow a = 2$
$\Rightarrow d = 4 \times 2 \left[ \text{ From } \left( 2 \right) \right]$
$\Rightarrow d = 8$

Thus, the A.P. is 2, 10, 18, 26, ..... .

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 5 Arithmetic Progression
Exercise 5.6 | Q 31 | Page 52