the sum of first 11 terms of an arithmetic progression
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use sn formulae for thi
Ravi062:
clear explanation please
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1
Hey,sup!
The data is missing in the question.
So I am assuming that the sum of first 11 natural numbers is asked.
As per the question,
a=1.
d=1.
n=11.
S(n)=n/2{ 2a+(n-1)d}.
=>S(11)=11/2{2*1+(11-1)1}
=11/2( 2+10).
= 11/2 × 12.
= 11×6.
= 66.
Hope it helps.
The data is missing in the question.
So I am assuming that the sum of first 11 natural numbers is asked.
As per the question,
a=1.
d=1.
n=11.
S(n)=n/2{ 2a+(n-1)d}.
=>S(11)=11/2{2*1+(11-1)1}
=11/2( 2+10).
= 11/2 × 12.
= 11×6.
= 66.
Hope it helps.
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