Question

find the sum of 9 term of A.P 8,6,4,......​

Answer 1

Answer:

Step-by-step explanation:

Let the number of terms is N.

Given,

First term ( a ) = -8

Common difference ( d ) = Difference of two consecutive terms

                               = ( -6 ) - ( -8 ) = -6 + 8 = 2.

Now,

⇒ Sum of  N terms of an A.P. = N/2 { 2 a + ( N - 1 ) d }

⇒ 52 = N/2 { 2 × ( -8 ) + ( N - 1 )2 }

⇒ 52 = N/2 { -16 + 2N - 2 }

⇒ 52 = N/2 ( -18 + 2N )

⇒ 52 × 2 = N ( -18 + 2N )

⇒ 104 = -18N + 2N²

⇒ 104 = 2 ( -9N + N² )

⇒ 104/2 = -9N + N²

⇒ 52 = -9N + N²

⇒ 0 = N² - 9N - 52

⇒ 0 = N² - 13N + 4N - 52

⇒ 0 = N ( N - 13 ) + 4 ( N - 13 )

⇒ 0 = ( N - 13 ) ( N + 4 )

∴ N = either 13 or -4.

But the numbers of terms in an A.P. can't be negative.

Therefore, the possible value is 13.

Hope it helps !!

Answer 2

Answer: For AP nth term is a+(n-1) d

here given

A1=8; A2 = 6

A3=4

Easy method by using formula

since Common Difference (d) = a2-a1

6-8=-2

Sum of 9th term for AP is

=n/2 [2a+(n-l) d ]

=9/2 [2(8)+(9-1) (-2)]

=9/2 [16-16]

=9/2 [o]

=0

So Sum of 9th term is zero.