Sum
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
Concept: Sum of First n Terms of an AP
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