Question

Find the sum of the first 40 positive integers divisible by 6.​

Answer 1

Answer:-

numbers divisible by 6 are 6,12,18,24.......

so, a=6,d=6,n=40

wkt,, sn=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..

Answer 2

We will add all the multiple of 6 till 40 which are :-

6,12,18,24,36,42,48,54,60,66,72,78,84,90,96,102,108,114,120,126,132,138,144,150,156,162,168,174,180,186,192,198,204,210,216,222,228,234

Now we will add them alll

6+12+18+24+36+42+48+54+60+66+72+78+84+90+96+102+108+114+120+126+132+138+144+150+156+162+168+174+180+186+192+198+204+210+216+222+228+234=4,650

Plz mark me the brainliest I have done a lot to find the answer