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Write first four terms of the AP, when the first term a and the common difference d are given as follows:

(i) a = 10, d = 10

(ii) a = -2, d = 0

(iii) a = 4, d = -3

(iv) a = -1, d = 12\dfrac{1}{2}

(v) a = -1.25, d = -0.25

AP

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Answer

(i) First four terms of the A.P. are given by :

⇒ a, a + d, a + 2d, a + 3d.

Substituting values we get :

⇒ 10, 10 + 10, 10 + 2 × 10, 10 + 3 × 10.

⇒ 10, 20, 30, 40.

Hence, first four terms of the A.P. are 10, 20, 30 and 40.

(ii) First four terms of the A.P. are given by :

⇒ a, a + d, a + 2d, a + 3d.

Substituting values we get :

⇒ -2, -2 + 0, -2 + 2 × 0, -2 + 3 × 0.

⇒ -2, -2, -2, -2.

Hence, first four terms of the A.P. are -2, -2, -2, -2.

(iii) First four terms of the A.P. are given by :

⇒ a, a + d, a + 2d, a + 3d.

Substituting values we get :

⇒ 4, 4 + (-3), 4 + 2 × (-3) + 4 + 3 × (-3).

⇒ 4, 4 - 3, 4 - 6, 4 - 9.

⇒ 4, 1, -2, -5.

Hence, first four terms of the A.P. are 4, 1, -2, -5.

(iv) First four terms of the A.P. are given by :

⇒ a, a + d, a + 2d, a + 3d.

Substituting values we get :

⇒ -1, 1+12,1+2×12,1+3×12-1 + \dfrac{1}{2}, -1 + 2 \times \dfrac{1}{2}, -1 + 3 \times \dfrac{1}{2}

⇒ -1, 2+12,1+1,1+32\dfrac{-2 + 1}{2}, -1 + 1, -1 + \dfrac{3}{2}

⇒ -1, 12,0,2+32-\dfrac{1}{2}, 0, \dfrac{-2 + 3}{2}

⇒ -1, 12,0,12-\dfrac{1}{2}, 0, \dfrac{1}{2}.

Hence, first four terms of the A.P. are 1,12,0,12-1, -\dfrac{1}{2}, 0, \dfrac{1}{2}.

(v) First four terms of the A.P. are given by :

⇒ a, a + d, a + 2d, a + 3d.

Substituting values we get :

⇒ -1.25, -1.25 + (-0.25) + -1.25 + 2 × -0.25, -1.25 + 3 × -0.25

⇒ -1.25, -1.25 - 0.25, -1.25 + (-0.5), -1.25 + (-0.75)

⇒ -1.25, -1.5, -1.75, -2

Hence, first four terms of the A.P. are -1.25, -1.5, -1.75, -2.

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