Question

Find the sum of the positive terms of the arithmetic sequence 85, 78, 71, ...​

Answer 1

Answer:

Given:

A.P = 85, 78, 71, . . . .

The first term of the A.P (a₁) is 85

Now,

78 - 85 = -7

71 - 78 = -7

Common difference is (d) = -7

By looking at the further numbers of the A.P, we get to know that (7) decreasing in each number, after some point negative number will arise,

We know that,

aₙ = a + (n-1)d

a₁₀ = 85 + (10 - 1)(-7)

a₁₀ = 85 + (9)(-7)

a₁₀ = 85 - 63

a₁₀ = 22

And,

a₁₅ = 85 + (15 - 1)(-7)

a₁₅ = 85 + (14)(-7)

a₁₅ = 85 - 98

a₁₅ = -13

And,

a₁₃ = 85 + (13 - 1)(-7)

a₁₃ = 85 + (12)(-7)

a₁₃ = 85 - 84

a₁₃ = 1

And,

a₁₄ = 85 + (14 - 1)(-7)

a₁₄ = 85 + (13)(-7)

a₁₄ = 85 - 91

a₁₄ = -6

Therefore,

a₁₃ = 1 and a₁₄ = -6

So, till a₁₃ = 1 positive terms are there,

We have to find the sum of 13 positive numbers.

Formula,

\boxed{\sf S_{n}=\frac{n}{2}[2a+(n-1)d]}

S

n

=

2

n

[2a+(n−1)d]

\sf S_{13}=\frac{13}{2}[2(85)+(13-1)(-7)]

Answer 2

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85, 78, 71,… is an A.P.

uₙ = a + (n-1) d

u₁ = a = 85

d = u₂ - u₁ = 78 - 85 = -7

Since we have to sum the positive terms,our terms must be greater than 0.

uₙ > 0

85 +(n-1) (-7) > 0

85 - 7n + 7 > 0

- 7n > - 92

7n < 92

n < 13.14

Thus, the first 13 terms are the positive terms.

Sₙ = n/2 {2a + (n-1)d}

S₁₃ = 13/2 {2(85) + 12(-7)}

S₁₃ = 13 { 85 -42 }

S₁₃ = 13 { 43 }

S₁₃ = 559