Sum of n terms of a certain series is given by S_n=2n+3n^2, what is the type of the series and what is its 20^(th) term?

2 Answers
Aug 9, 2017

It is an arithmetic progression with first term as 5 and common difference as 6 and 20^(th) term is 119

Explanation:

As sum of n terms of a certain series is given by S_n=2n+3n^2,

Sum of 20 terms is 2×20+3×20^2=40+1200=1240.

Further, sum of 19 terms is 2×19+3×19^2=38+1083=1121,.
Hence 20^(th) term is 1240-1121=119.

As sum of 1 term is 2×1+3×1^2=5, sum of first two terms is 2×2+3×2^2=4+12=16, second term is 16-5=11 and common difference is 11-5=6. If it is arithmetic progression the third term should be 11+6=17.

As sum of first three terms is 2×3+3×3^2=6+27=33, third term is 33-16=17 hence it is an arithmetic progression.

Aug 9, 2017

d=6,a_20=119,S_20=1240

Explanation:

"calculate the first 'few' terms of the sequence"

"using "S_n=2n+3n^2

S_1=2+3=5rArra_1=5

S_2=4+12=16

rArra_2=S_2-S_1=16-5=11

S_3=6+27=33

rArra_3=S_3-S_2=33-16=17

S_4=8+48=56

rArra_4=S_4-S_3=56-33=23

"the first 4 terms are "5,11,17,23

"common difference ( d)"

d=23-17=17-11=11-5=6

rArr" these terms are an arithmetic sequence with "d=6

"the sum to n terms of an arithmetic sequence is"

•color(white)(x)a_n=a_1+(n-1)d

rArra_20=5+(19xx6)=119

rArrS_20=(2xx20)+(3xx20^2)=1240