find the sum of first 25 terms of an AP in which 3rd and is 7 and 7th term is 2 more than thrice of its third term
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a3=a+2d
a7=2+3a3
Substitute value of a3
a+6d=2+3(a+2d)
a+6d=2+3a+6d
2a=2
a=1
a3=7
a+2d=7
2d=6
d=3
s25=25/2(2+24×3)
=25/2(74)
=900
a7=2+3a3
Substitute value of a3
a+6d=2+3(a+2d)
a+6d=2+3a+6d
2a=2
a=1
a3=7
a+2d=7
2d=6
d=3
s25=25/2(2+24×3)
=25/2(74)
=900
purvi328519:
the last answer is wiring
it is 925 not 900
wrong
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1
I hope I solved your problem so please mark my answer as brainly
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