Question

the sum of n terms of the series 2,5,8,11 is..60100 ,then n is?

Answer 1

2,5,8,11,..... is forming an AP
hence sum for an AP
$\frac{n}{2} (2a + (n - 1)d) \\ d = 3 \\ a = 2 \\we \: have \: to \: find \: n \\ \frac{n}{2} (4 + 3n - 3) = 60100 \\ \\ n + 3 {n}^{2} - 120200 = 0 \\ solving \: this \: you \: will \: get \\ n = 200$

Answer 2

Given AS = 2 , 5 , 8 , 11




First term ( a ) = 2



Common Difference = 4th term - 3rd term = 3rd term - 2nd term = 2nd term - 1st term

Common Difference ( d ) = 11 - 8 = 8 - 5 = 5 - 2 = 3



nth term = a + ( n - 1 )d

$\bold{a_{n} = }$ 2 + ( n - 1 )3

$\bold{a_{n} = }$  2 + 3n - 3

$\bold{a_{n} = }$ 3n - 1




We know that the sum of x terms from  1st term is $\dfrac{x}{2} [ a + a_{x} ]$



Therefore,


$S_{n} = \dfrac{n}{2}[ 2 + 3n - 1 ]$

$S_{n} = \dfrac{n}{2}[3n + 1]$


Hence, sum of n terms is $\dfrac{n}{2}[ 3n + 1 ]$



But according to the question, sum of n terms is 60100.

comparing $\dfrac{n}{2}[ 3n + 1 ]$ with 60100



$\dfrac{n}{2}[ 3n + 1 ]$ = 60100


3n² + n = 120200

3n² + n - 120200

( 3n + 601 )( n - 200 ) = 0




Terms can't be negative as it is the number of terms. therefore n = 200


Value of n = 200