find the sum of first term 'n' series a,3a,5a
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5
a,3a,5a,.......
first term (A)=a
common difference (d)=3a-a=2a
now
sum of nth term(Sn)=n/2 {2A+(n-1)d}
=n/2 {2a+(n-1) 2a}
Sn=n/2(2an)=an^2
first term (A)=a
common difference (d)=3a-a=2a
now
sum of nth term(Sn)=n/2 {2A+(n-1)d}
=n/2 {2a+(n-1) 2a}
Sn=n/2(2an)=an^2
abhi178:
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Answered by
10
Answer:
Sn = n/2 [ 2A + (n + 1) d ]
where A = a, d = 3a - a = 2a
∴ Sn = n/2 [ 2A + (n - 1) d ]
= n/2 [ 2a + (n - 1) 2a ]
= n/2 [ 2a + 2na - 2a ]
= n/2 (2na) = n^2 a
Step-by-step explanation:
@GENIUS
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