ℎ ℎ (−1)7=10 __________

1.	ℎ ℎ (−1)7=10 __________
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»

Here,given

a=1

d=4−1=3

and,s

=287

Now,

(2A+(n−1)d)

⇒287=

(2×1+(n−1)3)

⇒287=

(2+3n−3)

⇒574=n(3n−1)

⇒574=3n

−n

⇒3n

−n−574=0

onsolvingthequadraticequatonusingformula

−b±

−4ac

Wegetn=14&

−41

[doesnotexist]

so,n=14

Now,

(a+1)

⇒287=

(1+x)

⇒574=14(1+x)

⇒(1+x)=

574

⇒1+x=41

⇒x=41−1

∴x=40

x=40isthesolution.

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