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Question
If an = 3 – 4n, show that a1, a2, a3,... form an AP. Also find S20.
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Solution
Given that, nth term of the series is
an = 3 – 4n ...(i)
Put n = 1,
a1 = 3 – 4(1)
= 3 – 4
= –1
Put n = 2,
a2 = 3 – 4(2)
= 3 – 8
= –5
Put n = 3,
a3 = 3 – 4(3)
= 3 – 12
= –9
Put n = 4,
a4 = 3 – 4(4)
= 3 – 16
= –13
So, the series becomes –1, –5, –9, –13,...
We see that,
a2 – a1
= –5 – (–1)
= –5 + 1
= –4
a3 – a2
= –9 – (–5)
= –9 + 5
= –4
a4 – a3
= –13 – (–9)
= –13 + 9
= –4
i.e., a2 – a1 = a3 – a2 = a4 – a3 = ... = –4
Since the each successive term of the series has the same difference.
So, it forms an AP.
We know that, sum of n terms of an AP,
Sn = `n/2[2a + (n - 1)d]`
∴ Sum of 20 terms of the AP,
S20 = `20/2[2(-1) + (20 - 1)(-4)]`
= 10[–2 + (19)(–4)]
= 10(–2 – 76)
= 10 × (–78)
= –780
Hence, the required sum of 20 terms i.e., S20 is –780.
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