Question

ℎ ℎ (−1)7=10 __________

Answer 1

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

Answer:

x=40 is your answer

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