Find three rational numbers between -1
Answers
Common difference of AP is 4.
(b) First term of AP is 5.
Explanation:
Given information,
Fifth term of an arithmetic sequence is 21 and it's ninth term is 37.
(a) what is it's common difference?
(b) What is it's first term?
Here,
\sf a_{5}a
5
= 21
\sf a_{9}a
9
= 37
a = ?
d = ?
⚘ Using formula of nth term ::
\bf{\dag}\:{\boxed{\tt{a_{n} = a + (n - 1)d}}}†
a
n
=a+(n−1)d
⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━
(a)
➨ \tt a_{5} = a + (5 - 1)da
5
=a+(5−1)d
➨ \tt 21 = a + 4d21=a+4d
➨ \tt a = 21 - 4d\qquad- (1)a=21−4d−(1)
Also,
➨ \tt a_{9} = a + (9 - 1)da
9
=a+(9−1)d
➨ \tt 37 = a + 8d37=a+8d
➨ \tt a = 37 - 8d\qquad- (2)a=37−8d−(2)
From (1) & (2) we get,
➨ \tt 21 - 4d = 37 - 8d21−4d=37−8d
➨ \tt 21 - 37 = - 8d + 4d21−37=−8d+4d
➨ \tt \cancel{-} 16 = \cancel{-} 4d
−
16=
−
4d
➨ \tt 4d = 164d=16
➨ \tt d = {\cancel{\dfrac{16}{4}}}d=
4
16
➨ d = 4
Hence, common difference (d) of AP is 4.
(b)
Put d = 4 in (1) we get,
➨ \tt a = 21 - (4\:\times\:4)a=21−(4×4)
➨ \tt a = 21 - 16a=21−16
➨ a = 5
Hence, first term (a) of AP is 5.
Verification:
➨ \tt a_{5} = a + (5 - 1)da
5
=a+(5−1)d
➨ \tt 21 = a + 4d21=a+4d
By putting value of a and d in above equation we get,
➨ \tt 21 = 5 + (4\:\times\:4)21=5+(4×4)
➨ \tt 21 = 5 + 1621=5+16
➨ \tt 21 = 2121=21
➨ LHS = RHS
Also,
➨ \tt a_{9} = a + (9 - 1)da
9
=a+(9−1)d
➨ \tt 37 = a + 8d37=a+8d
By putting value of a and d in above equation we get,
➨ \tt 37 = 5 + (8\:\times\:4)37=5+(8×4)
➨ \tt 37 = 5 + 3237=5+32
➨ \tt 37 = 3737=37
➨ LHS = RHS
Hence, Verified ✔
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