The sum of the squares of the fifth and the eleventh term of an AP is

1.	The sum of the squares of the fifth and the eleventh term of an AP is 3 and the product of the second and the fourteenth term is equal it P. Find the product of the first and the fifteenth term of the AP. (a) (58P – 39)/45 (b) (98P + 39)/72 (c) (116P – 39)/90 (d) (98P + 39)/90
Answer» Correct option (c) (116P – 39)/90 Explanation: (a + 4d)² + (a + 10d)² = 3 → a² + 14ad + 58d² = 1.5. Also, (a + d)(a + 13d) = P → a² + 14ad + 13d² = P. Further, we need to find the value of a² + 14ad (product of the first and fifteenth terms of the AP). From the above two equations, we get that 45d² =3/2-p →13d² =(39-26p)/90 Thus, a² + 14ad = P −(39-26p)/90 = (116P − 39)/90.

The sum of the squares of the fifth and the eleventh term of an AP is 3 and the product of the second and the fourteenth term is equal it P. Find the product of the first and the fifteenth term of the AP. (a) (58P – 39)/45 (b) (98P + 39)/72 (c) (116P – 39)/90 (d) (98P + 39)/90

Answer»

Correct option (c) (116P – 39)/90

Explanation:

(a + 4d)² + (a + 10d)² = 3 → a² + 14ad + 58d² = 1.5.

Also, (a + d)(a + 13d) = P → a² + 14ad + 13d² = P. Further, we need to find the value of a² + 14ad (product of the first and fifteenth terms of the AP). From the above two equations, we get that 45d² =3/2-p →13d² =(39-26p)/90

Thus, a² + 14ad = P −(39-26p)/90 = (116P − 39)/90.

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The sum of the squares of the fifth and the eleventh term of an AP is 3 and the product of the second and the fourteenth term is equal it P. Find the product of the first and the fifteenth term of the AP. (a) (58P – 39)/45 (b) (98P + 39)/72 (c) (116P – 39)/90 (d) (98P + 39)/90

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