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6. Progressions
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Q3 of 105 Page 146

In an AP :

given a = 5, d = 3, an = 50 find n and sn

d is the common difference = 3

a is the first term = 5


an = 50


an = a + (n-1)d


50 = 5 + (n-1)(3)


45 = 3(n-1)


15 = n-1


n = 16


The sum of the series is Sn = [2a + (n-1) d]


= [ 2×5 + (16-1)(3)]


= 8 [ 55 ]


⇒ 440


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Questions · 105
6. Progressions
1 1 1 1 2 2 2 2 2 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 1 2 2 3 3 3 3 3 4 5 5 6 7 8 9 10 11 12 13 14 15 16 17 1 1 1 1 2 2 2 3 3 3 3 3 3 3 4 5 6 7 7 8 9 10 11 12 13 14 1 1 1 2 2 2 2 3 3 3 3 3 3 3 3 3 4 1 1 1 1 2 3 3 4 4 4 5 6 7
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