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Question
Show that a1, a2,..., an... form an AP where an is defined as below:
an = 9 − 5n
Also, find the sum of the first 15 terms.
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Solution
an = 9 − 5n
a1 = 9 − 5 × 1
= 9 − 5
= 4
a2 = 9 − 5 × 2
= 9 − 10
= −1
a3 = 9 − 5 × 3
= 9 − 15
= −6
a4 = 9 − 5 × 4
= 9 − 20
= −11
It can be observed that
a2 − a1 = −1 − 4 = −5
a3 − a2 = −6 − (−1) = −5
a4 − a3 = −11 − (−6) = −5
i.e., ak + 1 − ak is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.
`S_n = n/2 [2a + (n - 1)d]`
`S_15 = 15/2 [2(4) + (15 - 1) (-5)]`
= `15/2 [8 + 14(-5)]`
= `15/2 (8 - 70)`
= `15/2 (-62)`
= 15 × (-31)
= -465
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