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Question
The nth term of an AP is given by (−4n + 15). Find the sum of first 20 terms of this AP?
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Solution
Given, an = − 4n + 15
∴ a1 = − 4 × 1 + 15 = − 4 + 15 = 11
a2 = − 4 × 2 + 15 = − 8 + 15 = 7
a3 = − 4 × 3 + 15 = − 12 + 15 = 3
a4 = − 4 × 4 + 15 = − 16 + 15 = −1
It can be observed that
a2 − a1 = 7 − 11 = −4
a3 − a2 = 3 − 7 = −4
a4 − a3 = − 1 − 3 = −4
i.e., ak + 1 − ak is same every time. Therefore, this is an A.P. with common difference as
−4 and first term as 11.
`S_n=n/2[2a+(n-1)d]`
`S_20=20/2[2(11)+(20-1)(-4)]`
`=10[22+19(-4)]`
`=10(20-76)`
`=10(-54)`
`=-540`
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