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The sum of 6 and z how do solve this

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Answer:

let a and d be the first term and common

difference of ap

nth term of ap

a_{n} = a + (n - 1) * d

. a_{3} = a + (3 - 1) * (d = a + 2d)

aya+(7-1)da+6d

given a_{3} + a_{7} = 6

* (a + 2d) + (a + 6d) = 6

rightarrow 2a + 8 * d = 6

rightarrow a+4d=3.... (1)

also given

a_{3}*a_{7} = 8

therefore (a + 2d)(a + 6d) = 8

rightarrow (3 - 4d + 2d)(3 - 4d + 6d) = 8

[using

(1)]

(3-2d)(3+2d)=8

rightarrow 9 - 4d ^ 2 = 8 rightarrow 4d ^ 2 = 1

rightarrow d ^ 2 = 1/4

rightarrow d = plus/minus 1/2

when d = 1 2

a = 3 - 4 * d = 3 - 4 * 1/2 = 3 - 2 = 1

when d

1

2

a = 3 - 4 * (d = 3 + 4 * 1/2) = 3 + 2 = 5

when a=1\& d = 1/2

s_{16} = 16/2 * [2 * 1 + (16 - 1) * 1/2]

8(2 + 15/2) = 4 * 19 = 76

when a=5\& d = - 1/2

s_{16} = 16/2 * [2 * 5 + (16 - 1)(- 1/2)] =

8(10 - 15/2) = 4 * 5 = 20

thus, the sum of first 16 terms of the ap is 76 or 20.

User Mahdi Majidzadeh
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