Question

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

Answer 1

Answer:

✳ The AP is  2 , 10 , 18 , 26 , 34 , etc.

SOLUTION

GIVEN :-

1) The sum of first seven terms of an AP is 182.

2)4th and 17th terms are in the ratio 1:5.

TO FIND :-

The AP

ANSWER

Let us assume that the first term of the AP = a.

Let us assume that the common difference of the AP = d.

$\sf Sn = \dfrac{n}{2} (2a + (n - 1) * d)$

Here,

S₇ = 182

$\implies$

$\sf 182 = \dfrac{7}{2} (2a + (n - 1) * d)$

$\implies$

$\sf 182 = \dfrac{7}{2} (2a + 6d)$

$\implies$

$\sf 182 = 7*(a+3d)$

$\implies$

a + 3d = 182/7

$\implies$

a + 3d = 26 $\longrightarrow$(1)

We are also given that,

a₄ : a₁₇ = 1 : 5

$\sf \dfrac{a+3d}{a+16d} = \dfrac{1}{5}$

On Cross - Multiplication,

5a + 15d = a + 16d

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

$\implies$ 4a = d $\longrightarrow$ (2)

On Substituting d = 4a in (1),

a + 3 × 4a = 26

$\implies$ a + 12a = 26

$\implies$ 13a = 26

$\implies$

$\sf a = \dfrac{26}{13}$

$\implies$ First Term of the AP (a) = 2

Now, Let us substitute a = 2 in (2)

$\implies$ 4 × 2 = d

$\implies$ 8 = d

We know that an AP is of the form,

a, a + d, a + 2d, a + 3d, a + 4d

$\implies$ AP = 2 , 2 + 8 , 2 + (2×8) , 2 + (3×8) , etc .

$\implies$ AP = 2 , 10 , 18 , 26 , 34 , etc .