find the sum of 1,3,7..........20terms
Answers
Answer:
...
Step-by-step explanation:
We know that the general term of an arithmetic progression with first term a and common difference d is T
n
=a+(n−1)d
It is given that the 3rd term of the arithmetic series is 7 that is T
3
=7 and therefore,
T
3
=a+(3−1)d
⇒7=a+2d....(1)
Also it is given that the 7th term is 2 more than three times its 3rd term that is
T
7
=(3×T
3
)+2=(3×7)+2=21+2=23
Thus,
T
7
=a+(7−1)d
⇒23=a+6d....(2)
Subtract equation 1 from equation 2:
(a−a)+(6d−2d)=23−7
⇒4d=16
⇒d=
4
16
⇒d=4
Substitute the value of d in equation 1:
a+(2×4)=7
⇒a+8=7
⇒a=7−8=−1
We also know that the sum of an arithmetic series with first term a and common difference d is S
n
=
2
n
[2a+(n−1)d]
Now to find the sum of first 20 terms, substitute n=20,a=−1 and d=4 in S
n
=
2
n
[2a+(n−1)d] as follows:
S
20
=
2
20
[(2×−1)+(20−1)4]=10[−2+(19×4)]=10(−2+76)=10×74=740
Hence, the sum of first 20 terms is 740.
Step-by-step explanation:
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