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प्रश्न
If Sn denotes the sum of the first n terms of an A.P., prove that S30 = 3(S20 − S10)
उत्तर
Let a be the first term and d be the common difference.
We know that, sum of first n terms = Sn = \[\frac{n}{2}\] [2a + (n − 1)d]
Now,
S10 = \[\frac{10}{2}\] [2a + (10 − 1)d]
= 5(2a + 9d)
= 10a + 45d ....(1)
S20 = \[\frac{20}{2}\] [2a + (20 − 1)d]
= 10(2a + 19d)
= 20a + 190d ....(2)
S30 = \[\frac{30}{2}\] 302[2a + (30 − 1)d]
= 15(2a + 29d)
= 30a + 435d ....(3)
On subtracting (1) from (2), we get
S20 − S10 = 20a + 190d − (10a + 45d)
= 10a + 145d
On multiplying both sides by 3, we get
3(S20 − S10) = 3(10a + 145d)
= 30a + 435d
= S30 [From (3)]
Hence, S30 = 3(S20 − S10)
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