Question

the sum of four consecutive term of an ap is 32 and their product is 3465 find the terms ​

Answer 1

The  four consecutive term of an AP are 5, 7, 9 and 11.

Step-by-step explanation:

Let the  four consecutive term of an AP = (a - 3d), (a - d), (a + d), (a + 3d).

To find, the fourth terms of an AP = ?

According to question,

(a - 3d) + (a - d) + (a + d) + (a + 3d) = 32

⇒ 4a = 32

⇒  a = 8

Also,

(a - 3d)(a - d)(a + d)(a + 3d) = 32

⇒ $(a^{2} -d^2)(a^{2} -9d^2)=3465$

Put a = 8, we get

$(a^{2} -8^2)(a^{2} -9(8)^2)=3465$

⇒ $9d^4-640d^2+631=0$

⇒ $9d^4-631d^2-9d^2+631=0$

⇒ $9d^2(d^2-1)-631(d^2-1)=0$

⇒ $(9d^2-631)(d^2-1)=0$

⇒ $d^2-1=0$

⇒ d = 1

∴ The  four consecutive term of an AP = (8 - 3), (8 - 1), (8 + 1), (8 + 3)

i.e., 5, 7, 9 and 11

Thus, the  four consecutive term of an AP are 5, 7, 9 and 11.

Answer 2

Answer:

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