Question

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

Answer 1

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,

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

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

$\sf S_{13}=\frac{13}{2}[170+(12)(-7)]$

$\sf S_{13}=\frac{13}{2}[170-84]$

$\sf S_{13}=\frac{13}{2}\times 86$

$\sf S_{13}=13 \times 43$

$\sf S_{13}=559$

Hence,

The sum of the positive terms of the A.P is 559

Answer 2

Answer:

Given :-

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

To Find :-

Sum of positive terms

SoluTion :-

At first finding common Difference

78 - 85 = -7

78 - 71 = -7

Common Difference (D) = -7

Now,

We know that

aₙ = a + (n-1)d

a₁₀ = 85 + (10 - 1)(-7)

a₁₀ = 85 + (9)(-7)

a₁₀ = 85 - 9 × 7

a₁₀ = 85 - 63

a₁₀ = 22

Also,

a₁₃ = 85 + (13 - 1)(-7)

a₁₃ = 85 + (12)(-7)

a₁₃ = 85 + 12 × 7

a₁₃ = 85 - 84

a₁₃ = 1

Also,

a₁₄ = 85 + (14 - 1)(-7)

a₁₄ = 85 + (13)(-7)

a₁₄ = 85 + 13 × 7

a₁₄ = 85 - 91

a₁₄ = -6

Now,

Nth term = n/2 [2a(n - 1) d]

S₁₃ = 13/2[2(85)(13-1) (-7)]

S₁₃ = 13/2 [170(12)(-7)]

S ₁₃ = 13/2 [170 - 84]

S₁₃ = 13/2 [86]

S₁₃ = 559

The sum of the positive terms of the A.P is 559