In A.P if 1st S10=-80 & …
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Sia ? 6 years, 6 months ago
Let A.P be a , a + d, a + 2d, ...
a is the first term, d is the common difference
The formula of sum of n terms is
{tex}S_{n}=\frac{n}{2}[2 a+(n-1) d]{/tex}
Where n is the number of terms
The sum of its first 10 term is - 80. Therefore,
{tex}-80=\frac{10}{2}[2 a+(10-1) d]{/tex}
-80 = 5[2a + 9d]
-16 = 2a + 9d ...(1)
Sum of its next 10 terms is -280.
{tex}S_{20}=\frac{20}{2}[2 a+(20-1) d]{/tex}
{tex}S_{20}{/tex} = 10[2a + 19d]
Since sum of the next 10 terms of AP is -280. Therefore,
{tex}S_{20}-S_{10}{/tex} =-280
10[2a + 19d] - (-80) = -280
10[2a + 19d] = 360
2a + 19d = -36 ...(2)
Subtract (1) from (2)
2a + 19d - 2a - 9d = -36 + 16
10d = -20
d = -2
Put value of d in (1)
-16 = 2a + 9d
-16 = 2a + 9(-2)
-16 = 2a - 18
2 = 2a
a = 1
Therefore, The AP series is 1, -1, -3, -5 ...
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