Write the algebraic sequence of 1,8,15,21
Answers
G⃖I⃖V⃖E⃖N⃖ A⃖R⃖I⃖T⃖H⃖M⃖E⃖T⃖I⃖C⃖ S⃖E⃖Q⃖U⃖E⃖N⃖C⃖E⃖ I⃖S⃖ 9⃖, 1⃖5⃖, 2⃖1⃖,.
F⃖I⃖R⃖S⃖T⃖ T⃖E⃖R⃖M⃖ = A⃖ = 9⃖
C⃖O⃖M⃖M⃖O⃖N⃖ D⃖I⃖F⃖F⃖E⃖R⃖E⃖N⃖C⃖E⃖ = D⃖ =1⃖5⃖ 9⃖ = 6⃖
(A⃖) F⃖O⃖R⃖ N⃖A⃖T⃖U⃖R⃖A⃖L⃖ N⃖U⃖M⃖B⃖E⃖R⃖ N⃖
N⃖T⃖H⃖ T⃖E⃖R⃖M⃖ = [(N⃖−1⃖)X⃖6⃖]+9⃖=6⃖N⃖+3⃖ W⃖H⃖E⃖R⃖E⃖ N⃖=1⃖,2⃖,3⃖,
H⃖E⃖N⃖C⃖E⃖ A⃖L⃖G⃖E⃖B⃖R⃖A⃖I⃖C⃖ F⃖O⃖R⃖M⃖ O⃖F⃖ T⃖H⃖E⃖ S⃖E⃖Q⃖U⃖E⃖N⃖C⃖E⃖ 9⃖,1⃖5⃖,2⃖1⃖,. I⃖S⃖ X⃖
N⃖
=6⃖N⃖+3⃖.
(B⃖) N⃖T⃖H⃖ T⃖E⃖R⃖M⃖ =6⃖N⃖+3⃖
2⃖5⃖
T⃖H⃖
T⃖E⃖R⃖M⃖ =6⃖(2⃖5⃖)+3⃖=1⃖5⃖0⃖+3⃖=1⃖5⃖3⃖
(C⃖) S⃖U⃖M⃖ O⃖F⃖ F⃖I⃖R⃖S⃖T⃖ N⃖ T⃖E⃖R⃖M⃖S⃖ I⃖S⃖ G⃖I⃖V⃖E⃖N⃖ B⃖Y⃖,
S⃖
N⃖
=
2⃖
1⃖
N⃖[X⃖
1⃖
+X⃖
N⃖
]
S⃖U⃖M⃖ O⃖F⃖ 2⃖4⃖ T⃖E⃖R⃖M⃖S⃖ I⃖S⃖
S⃖
2⃖4⃖
=
2⃖
1⃖
N⃖[X⃖
1⃖
+X⃖
2⃖
4⃖]=
2⃖
2⃖5⃖
[9⃖+1⃖4⃖7⃖]=1⃖9⃖5⃖0⃖
S⃖U⃖M⃖ O⃖F⃖ 5⃖0⃖ T⃖E⃖R⃖M⃖S⃖ I⃖S⃖
S⃖
5⃖0⃖
=
2⃖
1⃖
N⃖[X⃖
1⃖
+X⃖
5⃖0⃖
]=
2⃖
5⃖0⃖
[9⃖+6⃖(5⃖0⃖)+3⃖]=7⃖8⃖0⃖0⃖
S⃖U⃖M⃖ O⃖F⃖ T⃖E⃖R⃖M⃖S⃖ F⃖R⃖O⃖M⃖ T⃖W⃖E⃖N⃖T⃖Y⃖ F⃖I⃖F⃖T⃖H⃖ T⃖O⃖ F⃖I⃖F⃖T⃖I⃖E⃖T⃖H⃖
=S⃖
5⃖0⃖
−S⃖
2⃖4⃖
=7⃖8⃖0⃖0⃖−1⃖9⃖5⃖0⃖
=5⃖8⃖5⃖0⃖
(D⃖) L⃖E⃖T⃖ S⃖U⃖M⃖ O⃖F⃖ F⃖I⃖R⃖S⃖T⃖ N⃖ T⃖E⃖R⃖M⃖S⃖ I⃖S⃖ 2⃖0⃖1⃖5⃖,
S⃖
N⃖
=
2⃖
1⃖
N⃖[X⃖
1⃖
+X⃖
N⃖
]
2⃖0⃖1⃖5⃖=
2⃖
1⃖
N⃖[9⃖+6⃖N⃖+3⃖]
2⃖0⃖1⃖5⃖=3⃖N⃖
2⃖
+6⃖N⃖
3⃖N⃖
2⃖
+6⃖N⃖−2⃖0⃖1⃖5⃖=0⃖
S⃖O⃖L⃖V⃖I⃖N⃖G⃖ A⃖B⃖O⃖V⃖E⃖ E⃖Q⃖U⃖A⃖T⃖I⃖O⃖N⃖ F⃖O⃖R⃖ N⃖, T⃖H⃖E⃖ V⃖A⃖L⃖U⃖E⃖ O⃖F⃖ N⃖ I⃖S⃖ N⃖O⃖T⃖ A⃖ N⃖A⃖T⃖U⃖R⃖A⃖L⃖ N⃖U⃖M⃖B⃖E⃖R⃖.
H⃖E⃖N⃖C⃖E⃖ 2⃖0⃖1⃖5⃖ C⃖A⃖N⃖ N⃖O⃖T⃖ B⃖E⃖ T⃖H⃖E⃖ S⃖U⃖M⃖ O⃖F⃖ S⃖O⃖M⃖E⃖ T⃖E⃖R⃖M⃖S⃖ O⃖F⃖ T⃖H⃖I⃖S⃖ S⃖E⃖Q⃖U⃖E⃖N⃖C⃖E⃖.
Step-by-step explanation:
Given arithmetic sequence is 9, 15, 21,.
First term = a = 9
Common difference = d =15 9 = 6
(A) For natural number n
nth term = [(n−1)x6]+9=6n+3 where n=1,2,3,
Hence algebraic form of the sequence 9,15,21,. is x
n
=6n+3.
(B) nth term =6n+3
25
th
term =6(25)+3=150+3=153
(C) Sum of first n terms is given by,
S
n
=
2
1
n[X
1
+X
n
]
Sum of 24 terms is
S
24
=
2
1
n[X
1
+X
2
4]=
2
25
[9+147]=1950
Sum of 50 terms is
S
50
=
2
1
n[X
1
+X
50
]=
2
50
[9+6(50)+3]=7800
Sum of terms from twenty fifth to fiftieth
=S
50
−S
24
=7800−1950
=5850