(iv) a=5 and d= 4 then find fifth term of an A.P
Section B
Answers
Answered by
0
Answer:
21
Step-by-step explanation:
the formula is a+(n-1)d
given,
t5= 5+(5-1)4
=16+5
=21
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Answered by
0
Answer:
ANSWER
(a)
x
5
40,x
10
=
20
x
10
−
x
5
5d
5d=
20−
40=
−20
⇒
d=
−4
Now,
x
15
=
x
10
+
5d=
20−
20=
0
(b) First term,
f=
x
5
4d=
40−
4(−4)=
56
S
n
2
n[2f+(n−1)d]
⇒
0=
2
n[2×56+(n−1)(−4)]
⇒
n(112−
4n+
4)=
0
⇒
n(116−
4n)=
0
⇒
=
0,n=
29
Thus,
29
terms of this sequence make the sum zero.
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