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Question
Show that a1, a2,..., an... form an AP where an is defined as below:
an = 3 + 4n
Also, find the sum of the first 15 terms.
Solution
an = 3 + 4n
a1 = 3 + 4(1) = 7
a2 = 3 + 4(2) = 3 + 8 = 11
a3 = 3 + 4(3) = 3 + 12 = 15
a4 = 3 + 4(4) = 3 + 16 = 19
It can be observed that
a2 − a1 = 11 − 7 = 4
a3 − a2 = 15 − 11 = 4
a4 − a3 = 19 − 15 = 4
i.e., ak + 1 − ak is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.
`S_n = n/2 [2a + (n - 1)d]`
`S_15 = 15/2 [2(7) + (15 - 1) × 4]`
= `15/2 [(14) + 56]`
= `15/2 (70)`
= 15 × 35
= 525
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