Question

solve the equation of AP 1+4+7+10+..........+X=287​

Answer 1

Answers

oshoraa

oshoraa Virtuoso

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

Answer 2

Answer:

Here,given

a=1

d=4−1=3

and,s

n

=287

Now,

s

n

=

2

n

(2a+(n−1)d)

⇒287=

2

n

(2×1+(n−1)3)

⇒287=

2

n

(2+3n−3)

⇒574=n(3n−1)

⇒574=3n

2

−n

⇒3n

2

−n−574=0

onsolvingthequadraticequatonusingformula

n=

2a

−b±

b

2

−4ac

Wegetn=14&

3

−41

[doesnotexist]

so,n=14

Now,

s

n

=

2

n

(a+1)

⇒287=

2

14

(1+x)

⇒574=14(1+x)

⇒(1+x)=

14

574

⇒1+x=41

⇒x=41−1

∴x=40

x=40 is the solution