The sum of the first 6 terms of an arithmetic progression is 138.
The 9th term and the 14th term are the ratio 9 : 13.
Find the first term, a1.
Answers
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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View Answer
If S
n
denotes the sum of the first terms of an AP, prove that S
30
=3(S
20
−S
10
Step-by-step explanation: