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Question
Show that a1, a2, a3, … form an A.P. where an is defined as an = 3 + 4n. Also find the sum of first 15 terms.
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Solution
an = 3 + 4n
a1 = 3 + 4 x 1 = 3 + 4 = 7
a2 = 3 + 4 x 2 = 3 + 8 = 11
a3 = 3 + 4 x 3 = 3 + 12 = 15
a4 = 3 + 4 x 4 = 3 + 16 = 19
and so on Here, a = 1 and d = 11 – 7 = 4
S15 = `n/(2)[2a + (n - 1)d]`
= `(15)/(2)[2 xx 7 + (15 - 1) xx 4]`
= `(15)/(2)[14 + 14 xx 14]`
= `(15)/(2)[14 + 56]`
= `(15)/(2) xx 70`
= 525.
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