Name: General Term of an AP: Example
Uploaded: 2019-07-19
Duration: 4 min 25 s
Description: Do you know how to find the first five terms of a sequence in A.P with the help of any two given terms of the sequence? Watch this video to learn how.

Question

The sum of four consecutive numbers in an

8

is

32

and the ratio of the product of the first and the last term to the product of two middle terms is

7 : 15

. Find the numbers.

Accepted Answer

Let the four consecutive number be $a, a + d, a + 2 d, a + 3 d$

Given, that sum of four consecutive numbers in $A . P$ is $32$ .

Therefore,

$a + a + d + a + 2 d + a + 3 d = 32 4 a + 6 d = 32 2 a + 3 d = 16 a = \frac{16 - 3 d}{2} . . . . . . . . . (1)$

The ratio of the product of the first and the last term to the product of two middle term is $7 : 15$ .

Therefore,

$\frac{a (a + 3 d)}{(a + d) (a + 2 d)} = \frac{7}{15} \Rightarrow 15 a (a + 3 d) = 7 (a + d) (a + 2 d) \Rightarrow 15 a^{2} + 45 a d = 7 a^{2} + 21 a d + 14 d^{2} \Rightarrow 8 a^{2} + 24 a d - 14 d^{2} = 0 \Rightarrow 4 a^{2} + 12 a d - 7 d^{2} = 0 \Rightarrow 4 a^{2} + 14 a d - 2 a d - 7 d^{2} = 0 \Rightarrow 2 a (2 a + 7 d) - d (2 a + 7 d) = 0 \Rightarrow (2 a + 7 d) (2 a - d) = 0 ∴ a = \frac{- 7 d}{2} & a = \frac{d}{2} . . . . . . . . . . . (2)$

Substitute the value of $a$ in equation $(2)$

So,

$a = \frac{- 7 d}{2} \Rightarrow \frac{16 - 3 d}{2} = \frac{- 7 d}{2} \Rightarrow - 4 d = 16 \Rightarrow d = - 4$

and

$a = \frac{d}{2} \Rightarrow \frac{16 - 3 d}{2} = \frac{d}{2} \Rightarrow 4 d = 16 \Rightarrow d = 4$

When $d = 4$

Then, $a = \frac{d}{2} = \frac{4}{2} = 2$

Therefore, four consecutive number in an $A P$ are

$2, (2 + 4), (2 + 2 \times 4), (2 + 3 \times 4) = 2, 6, 10, 14$

If we take $d = - 4$ then also we will get same numbers.

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