Question

2+5+8+11...+x=345
Solve this equation​

Answer 1

Step-by-step explanation:

2+5+8+11...+x=345. We observe the terms in the addition on left side are in Arithematic progression in which first term =a=2 and common difference = 3. Hence the sum of n terms in an Arithematic progression =

n/2(2a+(n-1)d ) Let's take the sum of n terms of this A.P. is 345 => n/2(2×2+(n-1)3) = 345 => n/2(4+3n-3) = 345 =>n/2(3n+1)=

345 => n(3n+1)=690 =>

$3 {n}^{2} + 46n - 45n - 690 = 0 = >$

$n(3n + 46) - 15(3n + 46) = 0 = >$

$= > (n - 15)(3n + 46) = 0 = >$

$(n - 15) = 0 \: or \: (3n + 46) = 0 = >$

$n = 15 \: or \: - \frac{46}{3} = >$

n can't be -46/3 because n can only be positive. Hence the sum of 15 terms in this A.P. =690. Hence the 15th term is x =>x=a+14d=2+14(3)=44. Answer is x=44. Hope it helps you.