If sum of 1st न terms of 3Ap are S1, S2, S3. The first term of each is 5 and their common difference are 2,4,6 respectively. Prove that S1 +S3 =2S2
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for S1
1st term,a=5
2nd term,a+d=7
3rd term,a+2d=9
n th term,a+(n-1)d=3+2n
so the ap is=5,7,9,.....
S1=n/2[5+3+2n]=n/2×[8+2n]=n^2+4n
for S2
1st term,a=5
2nd term=9
n th terma+(n-1)d=1+4n
S2=n/2[5+1+4n]=n/2[6+4n]=2n^2+3n
for S3
1st term=5
2nd term=11
n th term,a+(n-1)d=6n-1
S3=n/2[5+6n-1]=n/2[6n+4]=3n^2+2n
S1+S3=n^2+4n+3n^2+2n=4n^2+6n
=2×(2n^2+3n)=2S2
proved.
1st term,a=5
2nd term,a+d=7
3rd term,a+2d=9
n th term,a+(n-1)d=3+2n
so the ap is=5,7,9,.....
S1=n/2[5+3+2n]=n/2×[8+2n]=n^2+4n
for S2
1st term,a=5
2nd term=9
n th terma+(n-1)d=1+4n
S2=n/2[5+1+4n]=n/2[6+4n]=2n^2+3n
for S3
1st term=5
2nd term=11
n th term,a+(n-1)d=6n-1
S3=n/2[5+6n-1]=n/2[6n+4]=3n^2+2n
S1+S3=n^2+4n+3n^2+2n=4n^2+6n
=2×(2n^2+3n)=2S2
proved.
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