Math, asked by Hitha05, 1 year ago

if 1+4+7+10+..........+x = 287 ,find the value of x​

Answers

Answered by zaryaaab
3

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)

                                                                           -----------------------

                                                                                         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

Attachments:
Answered by rslekshmi08
0

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

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