Question

If for a sequence (tn),sn=2n2+5n find tn and show that the sequence is an a.p

Answer 1

Answer:

AP is 7,11,15...

Step-by-step explanation:

$S_n=2n^2+5n$

Substitute n = 1

$S_1=2(1)^2+5(1)$

$S_1=7$

So, First term is 7

Substitute n = 2

$S_2=2(2)^2+5(2)$

$S_2=18$

So, sum of first two terms is 18

Substitute n = 3

$S_3=2(3)^2+5(3)$

$S_3=33$

Sum of first 3 terms = 33

So, $t_3=S_3-S_2$

$t_3=33-18$

$t_3=15$

$t_2=S_2-S_1$

$t_2=18-7$

$t_2=11$

Common difference = $t_2-t_1 = t_3-t_2=11-7=15-11=4$

Since there exist a common difference

So, there exist an AP

So, AP is 7,11,15...