Find the sum of 30 terms of an AP in which 3Rd term is 11 and 10th term is one more than twice the 5th term
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a+2d= 11. ....................(1)
a+9d= 2(a+4d)+1
a-d= -1. ..........(2)
by equating 1 and 2
a= 3,d= 4
so sum= 30/2(2×3+29×4)
=15(6+116)
= 1830
a+9d= 2(a+4d)+1
a-d= -1. ..........(2)
by equating 1 and 2
a= 3,d= 4
so sum= 30/2(2×3+29×4)
=15(6+116)
= 1830
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