If a_1,a_2,a_3, ... are in A.P. such that a_1 + a_7 + a_16 = 40, then the sum of the first 15 terms of this A.P. is

[a_1, a_2, a_3, ... , a_n are in A.P.[a_1 + a_7 + a_16 = 40][a + (a + 6d) + (a + 15d) = 40][3a + 21d = 40 => a + 7d = \ (40 / 3)[15(a + 7d) = 200S_15 = (15 / 2) [ 2a + 14d ]] S_15 = 15(a + 7d)] [S_15 = 15 * (40 / 3) = 200

If a_{1},a_{2},a_{3}, are in A.P. such that a_{1} + a... | Sequence & Series (Mathematics)

$[a_1, a_2, a_3, \ldots, a_n$ { are in A.P.} $[a_1 + a_7 + a_{16} \\= 40][a + (a + 6d) + (a + 15d) = 40][3a + 21d$ = $40 \Rightarrow a + 7d$ = $\\ \frac{40}{3}[15(a + 7d) \\$ = $200S_15\\$ = $\frac{15}{2} \left[ 2a + 14d \right]]$ $S_15 = 15(a + 7d)] [S_{15}\\$ = 15 $\cdot \frac{40}{3} = 200$

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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Let \frac{1}{x_{1}},\frac{1}{x_{2}},,\frac{1}{x_{n}}( x_{i} eq 0 .\ for . \ i = 1,2,,n ) be in A.P. such that x_{1} = 4 and x_{21} = 20. If n is the least positive integer for which x_{n} > 50, then \sum_{i = 1}^{n}\mspace{2mu}( \frac{1}{x_{i}} ) is equal to

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