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Mathematics 23 Online
OpenStudy (anonymous):

find the sum of the first 15 numbers in a sequnce that has a1 = 6 and d = 4. a. 62 b.42 c.158 d.510

mathslover (mathslover):

Given Common Difference : d = 4 and the first term : \(a_1=6\)

mathslover (mathslover):

hence this is in arithmetic progression .... you must know the formula of sum of n terms of an arithmetic progression : \(\Huge{S_n=\frac{n}{2}({2a+(n-1)d})}\)

mathslover (mathslover):

hence \(\Huge{S_{15}=\frac{15}{2}(2(6)+(15-1)4)}\) \(\Huge{S_{15}=\frac{15}{2}(12+56)}\) \(\Huge{S_{15}=15*34}\) \(\Huge{S_{15}=510}\)

OpenStudy (anonymous):

thanks.

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