Question

if sum of n terms of A.P 2,5,8... is 155 then the number of terms is ?

Answer 1

first term, a = 2

common difference, d = 5-2 = 3

Sum of n terms of AP = 155

$\frac{n}{2} (2a + (n - 1)d) = 155 \\ \\ \Rightarrow \frac{n}{2} (2 \times 2 + (n - 1)3) = 155$

$\Rightarrow n (4+ 3n - 3) = 155 \times 2 \\ \\ \Rightarrow n(3n + 1) = 310$

$\Rightarrow 3 {n}^{2} + n - 310 = 0$

$n = \frac{ - 1 \pm \sqrt{ {1}^{2} - 4 \times 3 \times ( - 310) } }{2 \times 3}$

$n = \frac{ - 1 \pm \sqrt{ 1 + 3720 } }{6} \\ \\ n = \frac{ - 1 \pm \sqrt{ 3721 } }{6}$

$n = \frac{ - 1 \pm 61}{6} \\ \\ n = \frac{ - 62}{6} \: or \: \frac{60}{6}$

$n = \frac{60}{6} = 10$

n can't be -62/6 as n is a positive integer.

So n = 10

Number of terms of AP is 10.

Answer 2

Sn=155 a=2 d=3 Sn=n/2(2a+(n-1)d) Or,155=n/2(4+(n-1)3) Or,155=n/2(4+3n-3) Or,155=n/2(1+3n) Or,155=(3n^2+n)/2 Or,310=3n^2+n Or,3n^2+n-310=0 Or,3n^2-30n+31n-310=0 Or,3n(n-10)+31(n-10)=0 Or,(n-10)(3n+31)=0 So,n=10 as no.of terms cannot be negative.