Question

in an a .p if sum of its first n terms is 3n square +5n and its k term is 164 find the value of k

Answer 1

a=a1
thus S1=a1
3n2+5n, n=1
3(1)2 +5(1)=a1
3+5=a1
a=8
a2=S2-a
a2=3(2)2+5(2)-8
a2=12+10-8
a2=14
ak=a+(k-1)d
d=a2-a
d=14-8
d=6
ak=8+(k-1)6
164=8+(k-1)6
156=(k-1)6
k-1=26
k=27

Answer 2

$Sn \: = 3n {}^{2} + 5n \\ S1 \: = 3(1) {}^{2} + 5(1) \\ \: \: \: \: = 3 + 5 = 8 \\ S1 = a1 \\ 8 = a1 \\ S2 = 3(2) {}^{2} + 5(2) \\ = 12 + 10 = 22 \\ S2 = a1 + a2 \\ 22 = 8 + a2 \\ a2 = 22 - 8 \\ = 14 \\ ap \: is \: 8 \: 22 \: ..... \\ d = a2 - a1 = 14 - 8 = 6 \\ an = 164 \\ a + (n - 1)d = 164 \\ 8 + (n - 1)6 = 164 \\ 164 - 8 = (n - 1)6 \\ 156 \div 6 = n - 1 \\ 26 = n - 1 \\ n = 26 + 1 \\ = 27$
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