Question

please answer this question ​

Answer 1

Answer:

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=40isthesolution.

Answer 2

x = 40

Answer:

Here,given

a=1

d=4−1=3

and, sn = 287

Now,

sn = n/2(2a+(n−1)d)

⇒287= n/2(2×1+(n−1)3)

⇒287= n/2(2+3n−3)

⇒574=n(3n−1)

⇒574=3n^2−n

⇒3n^2−n−574=0

onsolvingthequadraticequatonusingformula

n= 2a−b± b2−4ac

We get n = 14 &41/-3 [doesnotexist]

so,n=14

Now, sn = 2n(a+1)

⇒287 = 214 (1+x)

⇒574=14(1+x)

⇒(1+x)= 574/14

⇒1+x=41

⇒x=41−1

∴x=40

x=40is the solution.

HOPE YOU GET YOUR SOLUTION⛄