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Mathematics

Assertion (A): The sum of first n terms of the A.P. −1, 5, 11, … is 3n2 − 4n.

Reason (R): The sum of first n terms of an A.P. is given by Sn = n2\dfrac{n}{2} [2a + (n − 1)d].

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

AP

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Answer

A.P. : -1, 5, 11, ……

Given,

a = -1

d = 5 - (-1) = 6

We know that,

Sn = n2\dfrac{n}{2} [2a + (n − 1)d]

⇒ Sn = n2\dfrac{n}{2} [2(-1) + (n − 1)6]

= n2\dfrac{n}{2} [-2 + (6n − 6)]

= n2\dfrac{n}{2} (6n − 8)

= n2×\dfrac{n}{2} \times 2(3n − 4)

= n(3n - 4)

= 3n2 - 4n.

∴ Assertion (A) is true.

The standard and correct formula for the sum of the first n terms of an A.P.

Sn = n2\dfrac{n}{2}[2a + (n − 1)d]

∴ Reason (R) is true.

Both A and R are true.

Hence, option 3 is the correct option.

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