Answers

2014-12-14T20:47:40+05:30
we know for arithmetic progression Sn= \frac{n}{2} [2a+(n-1)d]                                     Progression to be AP we should have
                    for value of n =1 we have S1 = 5
                                     n=2  we have  S2 =14
                                     n=3 we have  S3 = 27
                                     n=4 we have  S4 = 44
from above
     we have the numbers as 5, 9 , 13 , 17 and are in arithmetic progression
                                   we have  ⇒ d = 4
                                                     a = 5
     Nth term = a +(n-1)d= 5 +4n - 4= 4n + 1
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2014-12-15T13:24:00+05:30
Given that S_n = 2n^2 + 3n
S1 = 2 + 3 = 5
S2 = 8 + 6 = 14
S3 = 18 + 9 = 27
S4 = 32 + 12 = 44

t1 = 5,
t2 = S2 - S1 = 14 - 5 = 9
t3 = S3 - S2 = 27 - 14 = 13
t 4 = S4 - S3 = 44 - 27 = 17

The sequence is 5, 9 , 13, 17
This is an AP as the common difference is 4.

The nth term = tn = a + (n-1)d 
                             = 5 + (n - 1)4
                             = 5 + 4n - 4
                             = 4n + 1
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