Answers

2016-03-16T13:13:48+05:30
S_5 = \frac{5}{2}[2a_1+(5-1)d]
45* \frac{2}{5} = [2a_1+4d]
18 = 2[a_1+2d]
a_3 = 9                                                          ....(i)

S_{15} = \frac{15}{2}[2a_1+(15-1)d]
435* \frac{2}{15}= [2a_1+14d]
58 = 2[a_1+7d]
a_8=29                                                           ...(ii)

On solving from eq.(i)
a_1 + 2d = 9
a_1 = 9-2d                                                      ...(iii)

Putting a_1's value from eq.(iii) in eq.(ii) we get
a_8 = 29
a_1 + 7d = 29
9-2d+7d=29
5d = 20
d = 4                                                                ...(iv)

Putting d's value from eq.(iv) in eq.(ii) we get
a_1 = 9 - 2(4)
a_1 = 1                                                           ...(v)

Using eq.(iv) and eq.(v) we can get

S_n =  \frac{n}{2}[2a_1+(n-1)d]
S_n =  \frac{n}{2}[2(1)+(n-1)4]
S_n = n[1+2(n-1)]
S_n = n[1+2n-2]
S_n = n[2n-1]
S_n = 2n^2-n
1 1 1
2016-03-16T20:30:41+05:30
2n square - n is the sum of n terms
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