the sum of four consecutive terms of an ap is 32 and their product is 3465 find these terms
Answers
Step-by-step explanation:
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(a
2
−d
2
)(a
2
−9d
2
)=3465
Put a = 8, we get
(a^{2} -8^2)(a^{2} -9(8)^2)=3465(a
2
−8
2
)(a
2
−9(8)
2
)=3465
⇒ 9d^4-640d^2+631=09d
4
−640d
2
+631=0
⇒ 9d^4-631d^2-9d^2+631=09d
4
−631d
2
−9d
2
+631=0
⇒ 9d^2(d^2-1)-631(d^2-1)=09d
2
(d
2
−1)−631(d
2
−1)=0
⇒ (9d^2-631)(d^2-1)=0(9d
2
−631)(d
2
−1)=0
⇒ d^2-1=0d
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.