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In the given A.P.
3rd term is 7.
As per the given conditions, 7th term is (3×7)+2 = 23.
Let the 1st term be 'a' and common difference be 'd',
So, 3rd term = a + (3-1)×d = 7.....(1)
7th term = a + (7-1)×d = 23.....(2)
Subtract (1) from (2),
4×d = 16,
d = 4.
Put d = 4 in (1) to get a,
a = -1.
Now, Sum of first n terms is given by,
Sn = (n/2)[2a + (n-1)×d]
Sum of first 20 terms is,
S20 = (20/2)[2×(-1) + (20-1)×4] = 10 [-2 +80 - 4] = 10 × 74 = 740.
Therefore, Sum of first 20 terms is 740.
3rd term is 7.
As per the given conditions, 7th term is (3×7)+2 = 23.
Let the 1st term be 'a' and common difference be 'd',
So, 3rd term = a + (3-1)×d = 7.....(1)
7th term = a + (7-1)×d = 23.....(2)
Subtract (1) from (2),
4×d = 16,
d = 4.
Put d = 4 in (1) to get a,
a = -1.
Now, Sum of first n terms is given by,
Sn = (n/2)[2a + (n-1)×d]
Sum of first 20 terms is,
S20 = (20/2)[2×(-1) + (20-1)×4] = 10 [-2 +80 - 4] = 10 × 74 = 740.
Therefore, Sum of first 20 terms is 740.
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S=740.
It's a right answer.
It's a right answer.
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