ि Hu Lwn तप कि e Po\é,we Jﬂ_jdivisedle ﬁbé ' ‘ न

1.	ि Hu Lwn तप कि e Po\é,we Jﬂ_jdivisedle ﬁbé ' ‘ न
Answer» The first 40 positive integers divisible by 6 are 6,12,18,....... upto40 terms the given series is in arthimetic progression with first term a=6 and common difference d=6 sum of n terms of an A.p is n/2×{2a+(n-1)d} →required sum = 40/2×{2(6)+(39)6} =20{12+234} =20×246 =4920 no.s divisible by 6 are 6,12,18.......a=6d=6n=40Sn=(n/2)(2a+(n-1)d)Sn=(40/2)(26+(40-1)6)Sn=20(12+39*6)Sn=20(12+234)Sn=20(246)Sn=4920

Answer»

The first 40 positive integers divisible by 6 are 6,12,18,....... upto40 terms

the given series is in arthimetic progression with first term a=6 and common difference d=6

sum of n terms of an A.p is

n/2×{2a+(n-1)d}

→required sum = 40/2×{2(6)+(39)6}

=20{12+234}

=20×246

=4920

no.s divisible by 6 are 6,12,18.......a=6d=6n=40Sn=(n/2)(2a+(n-1)d)Sn=(40/2)(2*6+(40-1)*6)Sn=20(12+39*6)Sn=20(12+234)Sn=20(246)Sn=4920

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