Find the sum of the positive terms of the arithmetic sequence 85, 78, 71,
Answers
a=first term
d=common difference
here,a=85,d=-7
tn=a+(n-1)d
tn=85+(n-1)-7
tn=85+(7n+7)
92-7n=0
7n=92
n=92/7
n=13.14
n=13
Sn=(n/2)(2a+(n-1)d)
sn=13/2(2*85+(13-1)(-7)
sn=(170+(12*-7)
sn=170-84
sn=13/2*86
sn=559
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)]S
13
=
2
13
[2(85)+(13−1)(−7)]
\sf S_{13}=\frac{13}{2}[170+(12)(-7)]S
13
=
2
13
[170+(12)(−7)]
\sf S_{13}=\frac{13}{2}[170-84]S
13
=
2
13
[170−84]
\sf S_{13}=\frac{13}{2}\times 86S
13
=
2
13
×86
\sf S_{13}=13 \times 43S
13
=13×43
\sf S_{13}=559S
13
=559
Hence,
The sum of the positive terms of the A.P is 559