if 1+4+7+10+..........+x = 287 ,find the value of x
Answers
Answer:
40
Step-by-step explanation:
Sn = n/2(2a+(n-1)d) given a=1, d=4-1=3 & Sn = 287
287 = n/2 (2*1 +(n-1) 3)
287*2 = n(2 + 3n - 3)
574 = 2n + 3n^2 - 3n
3n^2 -n - 574 = 0
on solving the quadratic equation using formula n= -b + sq.root(b^2 -4ac)
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2a
we get n = 14, -41/3 n not equal to -41/3 due to negative nos.
n=14
Sn = n/2 (a +l)
287 = 14/2(1 +x)
574 = 14 (1+x)
574 / 14 = 1+x
41 = 1 + x
So, x = 41 - 1
x = 40 is the solution
this is an a.p with
a=1
d=3
S= 287
sum of n terms=(n/2)(2a+(n-1)d)
287= (n/2)(2+(n-1)3)
= (n/2)(2+3n-3)
= (n/2)(3n-1)
= (3n^2-n)÷2
574= 3n^2 -n
3n^2 - n - 574 = 0
from this n can be find out
then find the nth term