In an Ap sum of 4th and 8th term is 22 and product 2nd and 6th term is 33 find the AP
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Answered by
7
Answer:
4th term = first term +3d
8 th term = first term + 7d
(f+3d)+(f+7d) = 2f+10d
= 2(f+5d)
= 2×6th term=22
6th term = 22/2=11
6th term × 2nd term = 33
11×2nd term = 33
2nd term = 33/11= 3
6th term= f+5d
2nd term= f+ d
11-3= (f+5d)-(f+d)
8= 4d
d = 8/4=2
ap= 1,3,5,7,9,11,13,15,17,..............
Answered by
0
Answer:
Let a be the first term and d be the common difference of the given A.P. Then,
S
6
=42⟹
2
6
{2a+(6−1)d}=42⟹2a+5d=14 ...(i)
It is given that
a
10
:a
30
=1:3
⟹
a+29d
a+9d
=
3
1
⟹3a+27d=a+29d
⟹2a−2d=0
⟹a=d ...(ii)
putting the value of a in (i), we get
2d+5d=14⇒d=2
∴a=d=2
∴a
13
=a+12d=2+2×12=26
Hence, first term =2 and thirteenth term =26
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