Question

If in an AP, 8th term is 4 and 16th term is -20, then find the common difference of the given AP.​

Answer 1

Given :

• If in an AP, 8th term is 4 and 16th term is -20 .

To Find :

• Common Difference of the A.P .

Solution :

As Given that ,

$\longmapsto\tt{a+7d=4}$

$\longmapsto\tt{a+15d=-20}$

Now By Elimination Method :

$\longmapsto\tt{{\not{a}}+7d=4}$

$\longmapsto\tt{{\not{a}}+15d=-20}$

$\longmapsto\tt{-8d=24}$

$\longmapsto\tt{d=\cancel\dfrac{24}{-8}}$

$\longmapsto\tt\bf{d=-3}$

So , The Common Difference of the given A.P is -3 .

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• ${a}_{n}$=a + (n-1) × d
• General Form = a , a + d , a + 2d ...
• s = n/2 [2a (n-1) × d]

Here :

• n = nth term
• a = first term of the a.p
• s = sum of nth terms
• d = common difference

____________________

Answer 2

EXPLANATION.

In an A.P.

⇒ 8th term is 4.

⇒ 16th term is -20.

As we know that,

General term of an A.P.

⇒ Tₙ = a + (n - 1)d.

⇒ T₈ = a + (8 - 1)d.

⇒ T₈ = a + 7d.

⇒ a + 7d = 4. - - - - - (1).

⇒ T₁₆ = a + (16 - 1)d.

⇒ T₁₆ = a + 15d.

⇒ a + 15d = - 20. - - - - - (2).

From equation (1) and (2), we get.

Subtract both the equation, we get.

⇒ a + 7d = 4. - - - - - (1).

⇒ a + 15d = - 20. - - - - - (2).

⇒ -   -          +

We get,

⇒ - 8d = 24.

⇒ d = - 3.

Put the value of d = - 3 in the equation (1), we get.

⇒ a + 7d = 4.

⇒ a + 7(-3) = 4.

⇒ a - 21 = 4.

⇒ a = 4 + 21.

⇒ a = 25.

⇒ First term = a = 25.

⇒ Common difference = d = - 3.

Series = a, a + d, a + 2d, a + 3d, + . . . . .

25, (25 + (-3)), (25 + 2(-3)), (25 + 3(-3)), + . . . . .

⇒ Series = 25, 22, 19, 16, . . . . .

                                                                                                                       

MORE INFORMATION.

Supposition of terms in an A.P.

(1) = Three terms as : a - d, a, a + d.

(2) = Four terms as : a - 3d, a - d, a + d, a + 3d.

(3) = Five terms as : a - 2d, a - d, a, a + d, a + 2d.